Dual nature of Radiation and Matter - Class 12 Physics - Chapter 12 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Dual nature of Radiation and Matter | NCERT | Physics | Class 12
A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference $V$ esu. If $e$ and $m$ are the charge and mass of an electron, respectively, then the value of $h / \lambda$ (where $\lambda$ is the wavelength associated with the electron wave) is given by:
A) $2 \mathrm{meV}$
B) $\sqrt{\mathrm{meV}}$
C) $\sqrt{2 \mathrm{meV}}$
D) $\mathrm{meV}$
The correct answer to this problem is Option C: $\sqrt{2 \mathrm{meV}}$.
To understand why, let's use the de Broglie hypothesis, which relates the wavelength $\lambda$ of an electron to its momentum. The wavelength $\lambda$ of an electron can be expressed as:
$$ \lambda = \frac{h}{p} $$
Where:
- $h$ is Planck's constant.
- $p$ is the momentum of the electron.
For an electron accelerating through a potential difference $V$, its kinetic energy (KE) is given by:
$$ KE = eV $$
Here, $e$ is the electron's charge. The electron's kinetic energy also relates to its momentum $p$ through:
$$ KE = \frac{p^2}{2m} $$
Solving for $p$, we get:
$$ p = \sqrt{2mKE} = \sqrt{2meV} $$
Substituting back into the de Broglie equation for $\lambda$:
$$ \lambda = \frac{h}{\sqrt{2meV}} $$
Thus, the expression for $h / \lambda$ becomes:
$$ \frac{h}{\lambda} = \sqrt{2meV} $$
This derivation supports that the correct choice is Option C: $\sqrt{2 \mathrm{meV}}$.
A light ray behaves as:
A) a particle B) a wave C) both a particle and a wave D) a particle and a wave nature depending on the situation
The correct option is D) a particle and a wave nature depending on the situation.
Light exhibits dual nature characteristics, meaning that its behavior can be particle-like or wave-like, depending on the experimental conditions. For instance, light behaves as particles in phenomena such as the photoelectric effect, and demonstrates wave characteristics in processes like diffraction.
The $\mathrm{K}{\mathrm{a}}\mathrm{x}$-ray emission line of tungsten occurs at $\lambda = 21\ \mathrm{pm}$. What is the energy difference between $E{K}$ and $E_{\mathrm{L}}$ levels in this atom (MeV refers to mega electron volts; $1\ \mathrm{MeV} = 10^{6}\ \mathrm{eV}$ and $\left. \mathrm{hc} = 1242\ \mathrm{eV\ nm} \right)$?
A $\ 0.59\ \mathrm{MeV}$
B $\ 1.2\ \mathrm{MeV}$
C $\ 59\ \mathrm{KeV}$
D $\ 13.6\ \mathrm{eV}$
The correct answer is option A: $0.59$ MeV.
The K$\alpha$ X-ray emission in tungsten arises when an electron from the L-shell transitions to the K-shell, filling a vacancy. This shift in electron position between shells leads to the emission of an X-ray photon, the energy of which can be calculated by the difference in energy levels between the K-shell and L-shell, namely $\Delta E = E_K - E_L$.
Mathematically, the energy of the emitted photon, and hence the energy difference between the two energy levels, can be derived from the equation: $$ \Delta E = \frac{hc}{\lambda} $$
Inserting the given values (where $hc = 1242$ eV·nm, and $\lambda = 21$ pm or $0.021$ nm): $$ \Delta E = \frac{1242 \text{ eV·nm}}{0.021 \text{ nm}} = 591428.57 \text{ eV} $$
Converting eV to MeV (since $1$ MeV = $10^6$ eV): $$ E_K - E_L = 0.591 \text{ MeV} \approx 0.59 \text{ MeV} $$
Therefore, the energy difference between the K and L shells in the tungsten atom is 0.59 MeV.
Out of alpha ($\alpha$), beta ($\beta$), and gamma ($\gamma$) radiations, the radiations that are negatively charged are:
A) Both $\alpha$ and $\beta$ radiation
B) $\gamma$-radiation
C) Beta radiation
D) $\alpha$-radiation
Correct Answer: C) Beta radiation
Explanation: Beta radiation $(\beta)$ consists of negatively charged particles, specifically electrons. These electrons are characterized by their small mass and high velocity, allowing them to have significant penetrating power compared to alpha particles $(\alpha)$, which are positively charged. On the other hand, gamma rays $(\gamma)$ are neutral, as they are electromagnetic radiation and do not possess a charge. Therefore, among $\alpha$, $\beta$, and $\gamma$ radiations, only beta radiation is negatively charged.
A proton and an alpha particle are accelerated through a potential difference of $100 \mathrm{~V}$. The ratio of the wavelength associated with the proton to that associated with an alpha particle is
A. $\sqrt{2}: 1$
B. $2: 1$
C. $2 \sqrt{2}: 1$
D. $\frac{1}{2 \sqrt{2}}: 1$
The correct answer is C.
The wavelengths of particles like protons and alpha particles when they are accelerated through a potential difference can be calculated using the de Broglie wavelength formula: $$ \lambda = \frac{h}{\sqrt{2mqV}} $$ where $h$ is Planck's constant, $m$ is the mass of the particle, $q$ is the charge of the particle, and $V$ is the accelerating potential difference.
Given that the formula is proportional to: $$ \lambda \propto \frac{1}{\sqrt{mq}} $$ we can find the ratio of the wavelengths of a proton and an alpha particle: $$ \frac{\lambda_p}{\lambda_a} = \sqrt{\frac{m_a q_a}{m_p q_p}} $$ Considering that an alpha particle ($\alpha$) has twice the charge (2$e$) and four times the mass (4$m_p$) of a proton ($p$) ($m_p$ and $e$ being the mass and charge of a proton respectively): $$ \frac{\lambda_p}{\lambda_a} = \sqrt{\frac{4m_p \times 2e}{m_p \times e}} = \sqrt{8} = 2\sqrt{2} $$
The ratio of the wavelength associated with the proton to that associated with an alpha particle is $2\sqrt{2}:1$.
"Sir, what is radiation? (Briefly explain it)"
Radiation refers to the process of energy traveling through space. Common examples include light, heat, microwaves, and wireless communications, all of which represent different forms of radiation. Essentially, radiation can be described as the transfer of energy either through waves or fast-moving particles. One of the most well-known forms of radiation is sunshine.
An X-ray tube operates at $40 , \mathrm{kV}$. Suppose the electron converts $70%$ of its energy into a photon at each collision. Find the lowest three wavelengths emitted from the tube. Neglect the energy imparted to the atom with which the electron collides.
Solution:
Given Data:
- Voltage $V = 40 , \mathrm{kV} = 40 \times 10^3 , \mathrm{V}$
The energy of each electron is: $$ E = 40 \times 10^3 , \mathrm{eV} $$
Energy Utilized for Photon Generation (70% of total energy): $$ E_1 = \frac{70}{100} \times 40 \times 10^3 \mathrm{eV} = 28 \times 10^3 , \mathrm{eV} $$
Using the formula to find the wavelength $\lambda$ where $hc = 1242 , \mathrm{eV,nm}$, $$ \lambda_1 = \frac{hc}{E_1} = \frac{1242 , \mathrm{eV,nm}}{28 \times 10^3 , \mathrm{eV}} \approx 44.35 , \mathrm{nm} $$ The calculation yields a wavelength of $44.35 , \mathrm{pm}$ (pico-meters).
For the Second Wavelength:
- Using leftover energy (70% of the remaining energy): $$ E_2 = \frac{70}{100} \times (40 \times 10^3 - 28 \times 10^3) = \frac{70}{100} \times 12 \times 10^3 = 8.4 \times 10^3 , \mathrm{eV} $$
- Converting this energy into a wavelength, $$ \lambda_2 = \frac{1242}{8.4 \times 10^3} = 147.86 , \mathrm{pm} $$
For the Third Wavelength:
- Continuing this pattern, $$ E_3 = \frac{70}{100} \times (12 \times 10^3 - 8.4 \times 10^3) = 2.52 \times 10^3 , \mathrm{eV} $$
- Corresponding wavelength, $$ \lambda_3 = \frac{1242}{2.52 \times 10^3} \approx 493 , \mathrm{pm} $$
Summary of Results:
- First Wavelength: $\lambda_1 \approx 44.35 , \mathrm{pm}$
- Second Wavelength: $\lambda_2 \approx 148 , \mathrm{pm}$
- Third Wavelength: $\lambda_3 \approx 493 , \mathrm{pm}$
These calculations provide the lowest three wavelengths emitted by the X-ray tube under the specified conditions.
An X-ray tube produces a continuous spectrum of radiation with its short-wavelength end at 0.33 angstrom. What is the maximum energy of a photon in the radiation? Given Planck's constant $=6.6 \times 10^{-34} \mathrm{Js}$.
A) $35 \mathrm{keV}$
B) $37.5 \mathrm{keV}$
C) $40 \mathrm{keV}$
D) $42.5 \mathrm{keV}$
Solution
The correct answer is Option B: 37.5 keV.
To find the maximum energy of a photon in the radiation, we use the formula: $$ E_{\max} = \frac{h c}{\lambda_{\min }} $$ where:
- $h$ is Planck's constant ($6.6 \times 10^{-34} \mathrm{Js}$),
- $c$ is the speed of light ($3 \times 10^{8} \mathrm{m/s}$),
- $\lambda_{\min}$ is the shortest wavelength ($0.33$ angstroms, which is $0.33 \times 10^{-10}$ meters).
Substituting the given values: $$ E_{\max} = \frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{0.33 \times 10^{-10}} = 6 \times 10^{-15} \mathrm{J} $$ To convert this to keV, knowing $1 \mathrm{J} = 6.25 \times 10^{15} \mathrm{keV}$: $$ E_{\max} = 6 \times 10^{-15} \mathrm{J} \times 6.25 \times 10^{15} \mathrm{keV/J} = 37.5 \mathrm{keV} $$
Thus, the maximum energy of a photon is 37.5 keV, so the correct choice is (B) 37.5 keV.
"Formation of covalent bond between the atoms involved: a) Wave nature of electrons b) Particle nature of electrons c) Both a and b d) None"
The formation of a covalent bond between atoms involves the interaction of electrons, which are crucial components in atomic structures. According to Molecular Orbital Theory, atoms merge to form molecular orbitals, wherein the electrons are not confined to a single atom but shared between atoms across orbital spaces.
Electrons exhibit dual characteristics: they can be viewed as both particles and waves, reflecting their quantum mechanical nature. In the context of bonding, understanding both the wave and particle nature of electrons is essential. Therefore, the correct answer is:
- c) Both a and b
Order of $\left[\frac{\mathrm{e}}{\mathrm{m}}\right]$ ratio of proton, a particle, and electron is
A) $e>p>a$
B) $p>a>e$
C) $e>a>p$
D) None of these
The correct answer is Option A, which states the order as: $$ e > p > a $$
To understand this, we need to evaluate the charge-to-mass ratio ($\left[\frac{e}{m}\right]$) for each of these particles: the electron (e), the proton (p), and a generic particle (a). The ratios are calculated as follows:
-
For the proton: $$ \left[\frac{e}{m}\right]_p = \frac{e}{m_p} $$ where $e$ is the elementary charge and $m_p$ is the mass of a proton.
-
For particle 'a', assuming it represents an alpha particle (commonly used in physics problems): $$ \left[\frac{e}{m}\right]_a = \frac{2e}{4m_p} = \frac{e}{2m_p} $$ Note that an alpha particle has twice the charge of a proton but four times its mass.
-
For the electron: $$ \left[\frac{e}{m}\right]_e = \left[\frac{e}{m_e}\right] $$ Here, $m_e$ refers to the mass of an electron.
Based on these calculations, the electron, having a smaller mass compared to its charge, has the highest $\frac{e}{m}$ ratio. The proton, although having the same charge as the electron but a much higher mass, ranks second. The alpha particle, with double the charge but quadruple the mass of a proton, has the smallest ratio.
Thus, the order of $\frac{e}{m}$ values is: $$ \frac{e}{m_e} > \frac{e}{m_p} > \frac{e}{2m_p} $$ which confirms the order as electron (e) > proton (p) > alpha particle (a), corresponding to Option A: e > p > a.
A photocopier is a machine that makes paper copies of documents and other visual images quickly and cheaply. Most current photocopiers use a technology in which toner particles are positively charged and then transfer them onto paper in the form of an image/letters. What charge should the paper possess for this process to happen?
A. Only Positive
B. Only Negative
C. Only Neutral
D. Both B and C
The correct answer is D. Both B and C.
In this copying process, the toner particles are positively charged. For the positively charged toner to adhere well to the paper, the paper should ideally have a negative charge. Surprisingly, the process also works when the paper is neutral. This happens due to the principle of electrical induction. When positively charged toner particles are brought close to a neutral paper, the charges within the paper rearrange. This rearrangement induces a negative charge on the surface of the paper closer to the toner, causing the toner to stick due to the attraction between opposite charges.
Thus, whether the paper is negatively charged or neutral, it can successfully attract and hold the positively charged toner particles, making option D (Both B and C) the correct choice.
${ }^{1} \mathrm{C}$ undergoes $\beta^{+}$ decay to ${ }^{11} \mathrm{B}$. What is the maximum energy a positron emitted in the process can possess? Atomic masses: ${ }^{11} \mathrm{C}: 11.0114 \mathrm{u} ; { }^{11} \mathrm{B}: 11.0093 \mathrm{u} ; \mathrm{e}^{-}: 0.005486 \mathrm{u}; 931 \frac{\mathrm{MeV}}{\mathrm{u}}$.
A) Can't be uniquely determined. B) $1.44 \mathrm{MeV}$ C) $833.6 \mathrm{keV}$ D) $933.6 \mathrm{keV}$
Solution
The correct option is D) $933.6 , \text{keV}$
Let's first analyze the $\beta^{+}$ decay process: $$ { }{Z}^{A} X \rightarrow { }{Z-1}^{A} Y + e^{+} + v, $$ where $e^{+}$ represents a positron and $v$ is a neutrino.
The mass of the parent nucleus ${ }{Z}^{A} X$ is given by the atomic mass minus the mass of its electrons: $$ m({ }{Z}^{A} X) - Z \cdot m_{e}, $$ where $m_e$ is the electron (or positron) mass.
Similarly, the mass of the daughter nucleus ${ }{Z-1}^{A} Y$ is: $$ m({ }{Z-1}^{A} Y) - (Z-1) \cdot m_{e}. $$
Since the neutrino has negligible mass, the total mass of the products including the positron is: $$ m({ }{Z-1}^{A} Y) - (Z-1) \cdot m{e} + m_{e}. $$
The mass loss during the decay, which converts into kinetic energy (the $Q$-value), is: $$ \Delta m = m({ }{Z}^{A} X) - Z \cdot m{e} - [m({ }{Z-1}^{A} Y) - (Z-1) \cdot m{e} + m_{e}], $$ which simplifies to: $$ \Delta m = m({ }{Z}^{A} X) - m({ }{Z-1}^{A} Y) - 2 \cdot m_{e}. $$
Plugging in the given atomic masses for $^{11} \mathrm{C}$ and $^{11} \mathrm{B}$, and the mass of an electron: $$ \Delta m = (11.0114 , \mathrm{u} - 11.0093 , \mathrm{u} - 2 \times 0.005486 , \mathrm{u}) \times 931 \frac{\mathrm{MeV}}{\mathrm{u}}, $$ yields: $$ \Delta m = 0.933607 , \mathrm{MeV} = 933.6 , \mathrm{keV}. $$
Since the emitted energy is shared between the positron and the neutrino, the maximum energy that the positron can have is when the neutrino receives minimal to no energy. Thus, the maximum energy of the positron is $933.6 , \text{keV}$.
The energy of a photon of light with wavelength 5000 Å is approximately 2.5 eV. Therefore, the energy of an X-ray photon with wavelength 1 Å would be:
A) $\frac{2.5}{5000 \mathrm{eV}}$
B) $\frac{2.5}{(5000)^2 \mathrm{eV}}$
C) $2.5 \times 5000 \mathrm{eV}$
D) $2.5 \times (5000)^2 \mathrm{eV}$
The correct answer is Option C, given by:
$$ 2.5 \times 5000 , \text{eV} $$
This is derived based on the principle that photon energy ($E$) is inversely proportional to its wavelength ($\lambda$), i.e., $E \propto \frac{1}{\lambda}$. The formula used to compare energies at different wavelengths for photons can be distilled from this relationship:
$$ \frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1} $$
Given that a photon with wavelength $5000 , \text{Å}$ has energy $2.5 , \text{eV}$, and you are asked to find the energy of an X-ray photon with a wavelength of $1 , \text{Å}$, applying the formula:
$$ \frac{2.5 , \text{eV}}{E_{\text{X-ray}}} = \frac{1 , \text{Å}}{5000 , \text{Å}} $$
Thus, solving for $E_{\text{X-ray}}$:
$$ E_{\text{X-ray}} = 2.5 \times 5000 , \text{eV} = 12500 , \text{eV} $$
This matches Option C.
The photoelectric emission requires a threshold frequency $\vartheta_{\circ}$ for a certain metal. If incident photons have wavelengths $\lambda_{1}=2200 \text{ \AA}$ and $\lambda_{2}=1900$ Å producing two photoelectrons with $KE_{1}$ and $KE_{2}$ respectively, and if $KE_{1}=2KE_{2}$, then
(A) $\boldsymbol{\vartheta}_{0}$ is $1.1483 \times 10^{15} \text{ sec}^{-1}$
(B) $\lambda_{0}$ is $2.6126 \times 10^{-7} \text{ m}$
(C) $\boldsymbol{\lambda}_{0}$ is $1.2375 \times 10^{-7} \text{ m}$
(D) $\boldsymbol{\vartheta}_{0}$ is $2.16 \times 10^{-7} \text{ sec}^{-1}$
The correct answer is (C) $\lambda_{0} = 1.2375 \times 10^{-7} \text{ m}$. Here's the explanation of the problem:
Using the photoelectric effect equation for energy, the energy of the incident photon should be equal to the sum of the work function ($\phi = h \vartheta_0$) and the kinetic energy of the ejected electron: $$ \frac{hc}{\lambda} = h \vartheta_0 + KE $$
Given $KE_1 = 2 KE_2$, and if we denote $KE_1 = 2KE$ and $KE_2 = KE$, the equations using the wavelengths $\lambda_1$ and $\lambda_2$ can be written as: $$ \frac{hc}{\lambda_1} = h\vartheta_0 + 2KE $$ $$ \frac{hc}{\lambda_2} = h\vartheta_0 + KE $$
From these, we can express kinetic energies with respect to threshold frequency and the wavelengths: $$ 2KE = \frac{hc}{\lambda_2} - h\vartheta_0 $$ $$ \frac{hc}{\lambda_1} = \frac{hc}{\lambda_2} - h\vartheta_0 \implies 2\left( \frac{hc}{\lambda_1} - h\vartheta_0 \right) = \frac{hc}{\lambda_2} - h\vartheta_0 $$
Simplifying these equations to isolate $h\vartheta_0$ (and replacing $hc$ and $h\vartheta_0$ by common terms), we have: $$ 2\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_0}\right) = \frac{1}{\lambda_2} - \frac{1}{\lambda_0} $$
Expressing $\frac{1}{\lambda_0}$ in terms of $\frac{1}{\lambda_1}$ and $\frac{1}{\lambda_2}$: $$ \frac{1}{\lambda_0} = \frac{2}{\lambda_2} - \frac{1}{\lambda_1} $$
By solving and plugging in the wavelengths $2200 \text{ Å}$ for $\lambda_1$ and $1900 \text{ Å}$ for $\lambda_2$, one finds: $$ \frac{1}{\lambda_0}=\frac{\frac{2}{1900} - \frac{1}{2200}}{1} $$ $$ \lambda_0 = 1.2375 \times 10^{-7} \text{ m} $$
Therefore, option (C) is the right choice, stating that $\lambda_{0} = 1.2375 \times 10^{-7} \text{ m}$.
"Is light considered as matter?"
Light is an electromagnetic wave and represents a form of energy rather than matter. This distinction is crucial because matter is composed of particles with mass, whereas light is massless and does not fit the criteria that define matter.
The energy levels of a certain atom for the 1st, 2nd, and 3rd levels are $E, \frac{4E}{3}$, and $2E$ respectively. A photon of wavelength $\lambda$ is emitted for a transition from level 3 to 1. What will be the wavelength of emission for the transition from level 2 to 1?
A) $\frac{\lambda}{3}$
B) $\frac{4\lambda}{3}$
C) $\frac{3\lambda}{4}$
D) $3\lambda$
The correct option is D) $3\lambda$
To analyze these transitions between energy levels, we can use the equation for energy change during photon emission: $$ \Delta E = E_i - E_f = \frac{hc}{\lambda} $$ where $\Delta E$ is the energy difference between the initial ($E_i$) and final ($E_f$) states, $hc$ is the Planck's constant times the speed of light, and $\lambda$ is the wavelength of the emitted photon.
Given Data:
- Energy for level 1: $E$
- Energy for level 2: $\frac{4E}{3}$
- Energy for level 3: $2E$
Transition from level 3 to 1:
$$ \Delta E = E_3 - E_1 = 2E - E = E $$ Which gives: $$ E = \frac{hc}{\lambda} $$
Transition from level 2 to 1:
$$ \Delta E = E_2 - E_1 = \frac{4E}{3} - E = \frac{E}{3} $$ For the emitted photon during this transition, we get: $$ \frac{E}{3} = \frac{hc}{\lambda'} $$ Substituting $E$ from the first transition: $$ \frac{\frac{hc}{\lambda}}{3} = \frac{hc}{\lambda'} $$ This simplifies to: $$ \frac{hc}{3\lambda} = \frac{hc}{\lambda'} $$ Thus: $$ \lambda' = 3\lambda $$
Therefore, the wavelength of the emitted photon for the transition from level 2 to 1 is $3\lambda$, which corresponds to option D).
Two forces are in the ratio of 5:2. The maximum and minimum of their resultants are in the ratio
A. 5:2
B. 2:5
C. 7:3
D. 3:7
To solve the problem, let's break it down step-by-step.
Given:
Two forces are in the ratio of 5:2.
We need to find the ratio of their maximum and minimum resultants.
Step 1: Understanding Resultant Force
The resultant of two forces $ \mathbf{F_1} $ and $ \mathbf{F_2} $ is given by:
$$ \mathbf{R} = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta} $$
Where $ \theta $ is the angle between the two forces.
Step 2: Calculate the Maximum Resultant
The maximum value of the resultant occurs when $ \cos\theta = 1 $, which corresponds to $ \theta = 0^\circ $.
Given the forces are in a ratio of 5:2, we can let:
$$ F_1 = 5x \quad \text{and} \quad F_2 = 2x $$
So, the maximum resultant $ R_{max} $ is:
$$ R_{max} = \sqrt{(5x)^2 + (2x)^2 + 2 \cdot (5x) \cdot (2x) \cdot \cos(0^\circ)} $$ $$ R_{max} = \sqrt{25x^2 + 4x^2 + 20x^2} $$ $$ R_{max} = \sqrt{49x^2} $$ $$ R_{max} = 7x $$
Step 3: Calculate the Minimum Resultant
The minimum value of the resultant occurs when $ \cos\theta = -1 $, which corresponds to $ \theta = 180^\circ $.
So, the minimum resultant $ R_{min} $ is:
$$ R_{min} = \sqrt{(5x)^2 + (2x)^2 + 2 \cdot (5x) \cdot (2x) \cdot \cos(180^\circ)} $$ $$ R_{min} = \sqrt{25x^2 + 4x^2 - 20x^2} $$ $$ R_{min} = \sqrt{9x^2} $$ $$ R_{min} = 3x $$
Step 4: Ratio of Maximum to Minimum Resultants
The ratio of the maximum to minimum resultants is:
$$ \frac{R_{max}}{R_{min}} = \frac{7x}{3x} = \frac{7}{3} $$
Thus, the ratio of the maximum to the minimum resultants is 7:3.
Final Answer
Therefore, the answer is C. 7:3.
To demonstrate the phenomenon of interference, we require two sources which emit radiation of the same frequency, two sources which emit radiation of nearly the same frequency, and two sources which emit radiation of different wavelengths.
To demonstrate the phenomenon of interference, we require coherent sources.
What Are Coherent Sources?
Coherent sources are defined as sources that emit radiation of the same frequency and maintain a constant phase difference over time.
Conditions for Interference
Same Frequencies: For coherence, the frequencies of the two sources, say $ f_1 $ and $ f_2 $, must be equal. Hence, $ f_1 = f_2 = f $.
Definite Phase Relationship: The phase difference between the two sources must remain constant over time.
Given the options:
Nearly the same frequency
Same frequency
Different wavelengths
Same frequency and having a definite phase relationship
The most appropriate choice is: Option D: Same frequency and having a definite phase relationship.
This option ensures that the sources are coherent, satisfying both the frequency and phase relationship criteria needed for interference.
Therefore, the correct answer is: Option D
Two forces $\vec{F}_{1}$ and $\vec{F}_{2}$ are acting on a body. One force is double that of the other force, and the resultant is equal to the greater force. Then the angle between the two forces is:
To determine the angle between the two forces, let's analyze the given conditions step-by-step.
Given:
Two forces: $\vec{F}_{1}$ and $\vec{F}_{2}$
One force is double the other: Let's assume $\vec{F}_{2} = 2\vec{F}_{1}$
Resultant is equal to the greater force: The resultant $\vec{R}$ is equal to $\vec{F}_{2}$. Therefore, $\vec{R} = 2\vec{F}_{1}$
The resultant of the two forces can be expressed as: $$ R = \sqrt{ F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos \theta } $$
Substitute the values into the equation: $$ 2F_{1} = \sqrt{ F_{1}^{2} + (2F_{1})^{2} + 2 \cdot F_{1} \cdot 2F_{1} \cdot \cos \theta } $$
Simplify the equation: $$ 2F_{1} = \sqrt{ F_{1}^{2} + 4F_{1}^{2} + 4F_{1}^{2} \cos \theta } $$ $$ 2F_{1} = \sqrt{ 5F_{1}^{2} + 4F_{1}^{2} \cos \theta } $$
Square both sides to remove the square root: $$ (2F_{1})^{2} = 5F_{1}^{2} + 4F_{1}^{2} \cos \theta $$ $$ 4F_{1}^{2} = 5F_{1}^{2} + 4F_{1}^{2} \cos \theta $$ $$ 4F_{1}^{2} - 5F_{1}^{2} = 4F_{1}^{2} \cos \theta $$ $$ -F_{1}^{2} = 4F_{1}^{2} \cos \theta $$ $$ \cos \theta = -\frac{1}{4} $$
Thus, the angle $\theta$ between the two forces is: $$ \theta = \cos^{-1} \left( -\frac{1}{4} \right) $$
This evaluates to: $$ \theta = 135^{\circ} $$
Therefore, the angle between the two forces is $135^{\circ}$.
Two vectors $\vec{P}$ and $\vec{Q}$ are having their magnitudes in the ratio $\sqrt{3}: 1$. Further $|\vec{P} + \vec{Q}| = |\vec{P} - \vec{Q}|$ then the angle between the vectors $(\vec{P} + \vec{Q})$ and $(\vec{P} - \vec{Q})$ is
Given two vectors ( \vec{P} ) and ( \vec{Q} ) with magnitudes in the ratio ( \sqrt{3} : 1 ), we have the following conditions: [ |\vec{P}| = \sqrt{3} |\vec{Q}| ] Furthermore, it is provided that: [ |\vec{P} + \vec{Q}| = |\vec{P} - \vec{Q}| ]
Step-by-Step :
Square Both Sides: By squaring both sides of ( |\vec{P} + \vec{Q}| = |\vec{P} - \vec{Q}| ), we get: [ (\vec{P} + \vec{Q}) \cdot (\vec{P} + \vec{Q}) = (\vec{P} - \vec{Q}) \cdot (\vec{P} - \vec{Q}) ]
Expand Both Sides: Expand the dot products: [ \vec{P} \cdot \vec{P} + 2 \vec{P} \cdot \vec{Q} + \vec{Q} \cdot \vec{Q} = \vec{P} \cdot \vec{P} - 2 \vec{P} \cdot \vec{Q} + \vec{Q} \cdot \vec{Q} ]
Simplify: Simplifying the equation, we get: [ 2 \vec{P} \cdot \vec{Q} = -2 \vec{P} \cdot \vec{Q} ] This implies: [ 4 \vec{P} \cdot \vec{Q} = 0 \implies \vec{P} \cdot \vec{Q} = 0 ] Thus, the vectors ( \vec{P} ) and ( \vec{Q} ) are perpendicular to each other.
Angle Between ( \vec{P} ) and ( \vec{Q} ): Since ( \vec{P} \cdot \vec{Q} = 0 ), the angle ( \theta ) between ( \vec{P} ) and ( \vec{Q} ) is ( 90^\circ ).
Angle Between ( (\vec{P} + \vec{Q}) ) and ( (\vec{P} - \vec{Q}) ):
Given that ( \vec{P} ) and ( \vec{Q} ) are perpendicular to each other, the angle between ( \vec{P} + \vec{Q} ) and ( \vec{P} - \vec{Q} ) follows:
Using the formula for the dot product to find the angle between ( (\vec{P} + \vec{Q}) ) and ( (\vec{P} - \vec{Q}) ): [ (\vec{P} + \vec{Q}) \cdot (\vec{P} - \vec{Q}) = \vec{P} \cdot \vec{P} - \vec{Q} \cdot \vec{Q} = P^2 - Q^2 ] Since the magnitudes of ( \vec{P} ) and ( \vec{Q} ) are in the ratio ( \sqrt{3} : 1 ): [ |\vec{P}| = \sqrt{3} |\vec{Q}| \implies |\vec{P}|^2 = 3 |\vec{Q}|^2 ]
Substituting the magnitudes: [ = 3|\vec{Q}|^2 - |\vec{Q}|^2 = 2|\vec{Q}|^2 ] Since the dot product ( (\vec{P} + \vec{Q}) \cdot (\vec{P} - \vec{Q}) ) is positive, and it’s half the sum of squared magnitudes, the angle between these vectors is: [ \boxed{60^\circ} ]
Thus, the angle between the vectors ( (\vec{P} + \vec{Q}) ) and ( (\vec{P} - \vec{Q}) ) is ( 60^\circ ).
If h is Planck's constant and λ is the wavelength, $\frac{h}{λ}$ has the dimensions of:
A. Energy
B. Momentum
C. Moment of Inertia
D. Frequency
If h is Planck's constant and λ is the wavelength, $\frac{h}{λ}$ has the dimensions of:
A. Energy
B. Momentum
C. Moment of Inertia
D. Frequency
To determine the dimensions of $\frac{h}{\lambda}$, let's follow these steps:
Planck's constant ($h$) and wavelength ($\lambda$) are given.
The relationship between energy ($E$) and these quantities can be expressed as: $$ E = h \nu $$ Since frequency ($\nu$) can be written as $\nu = \frac{c}{\lambda}$ (where $c$ is the speed of light), we have: $$ E = h \frac{c}{\lambda} $$
Rearranging this formula, we get: $$ \frac{h}{\lambda} = \frac{E}{c} $$
This implies that $\frac{h}{\lambda}$ has the same dimensions as $\frac{E}{c}$.
Let's now analyze the dimensions of energy ($E$) and the speed of light ($c$):
Dimensions of energy ($E$):
$$ [E] = [M^1 L^2 T^{-2}] $$Dimensions of the speed of light ($c$):
$$ [c] = [M^0 L^1 T^{-1}] $$
Therefore, the dimensions of $\frac{E}{c}$ are: $$ \left[ \frac{E}{c} \right] = \frac{[M^1 L^2 T^{-2}]}{[M^0 L^1 T^{-1}]} = [M^1 L^{2-1} T^{-2+1}] = [M^1 L^1 T^{-1}] $$
These are precisely the dimensions of momentum.
Hence, $\frac{h}{\lambda}$ has the dimensions of momentum.
Final Answer: B. Momentum
The speed of a projectile at its maximum height is $\sqrt{3}/2$ times its initial speed. If the range of the projectile is $P$ times the maximum height attained by it, $P$ is equal to:
A $4/3$
B $2\sqrt{3}$
C $4\sqrt{3}$
D $3/4$
To determine the value of $ P $ (denoted as $ n $ in the transcript), we need to follow these steps:
Given Data: The speed of the projectile at its maximum height is $ \frac{\sqrt{3}}{2} $times its initial speed $ (u)$.
At the maximum height, the vertical component of the velocity $ v_y $ is zero. The horizontal component of the velocity remains constant throughout the motion and is given by $u \cos \theta$.
Therefore, at the maximum height: $$ \frac{\sqrt{3}}{2} u = u \cos \theta. $$ Simplifying this, we get: $$ \cos \theta = \frac{\sqrt{3}}{2}. $$ The angle $ \theta $ is $ 30^\circ $ because $ \cos 30^\circ = \frac{\sqrt{3}}{2} $.
Range and Maximum Height:
The range $ R $ of the projectile is given by: $$ R = \frac{u^2 \sin(2\theta)}{g} = \frac{u^2 \sin(60^\circ)}{g} = \frac{u^2 \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3}u^2}{2g}. $$
The maximum height ( H ) of the projectile is given by: $$ H = \frac{u^2 \sin^2(\theta)}{2g} = \frac{u^2 \sin^2(30^\circ)}{2g} = \frac{u^2 \left(\frac{1}{2}\right)^2}{2g} = \frac{u^2}{8g}. $$
Calculating ( P ):
We need $ P$ such that $R = P \cdot H $. Substituting the values of $R$ and $ H $: $$ \frac{\sqrt{3}u^2}{2g} = P \cdot \frac{u^2}{8g}. $$
Simplifying this equation, we get: $$ \sqrt{3}u^2 \cdot 8 = 2P \cdot u^2. $$
Canceling the common terms, we have: $$ 8\sqrt{3} = 2P. $$
Finally: $$ P = 4\sqrt{3}. $$
Therefore, the correct answer is C: $4\sqrt{3}$.
The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. The coefficient of friction when the angle of inclination of the plane is 60° is:
A $\frac{1}{3}$
B $\frac{1}{\sqrt{2}}$
C $\frac{1}{\sqrt{3}}$
D $\frac{1}{2}$
Let's break down the problem step-by-step.
Problem Statement
Given:
The force required to move a body up a rough inclined plane is twice the force required to prevent the body from sliding down the plane.
The angle of inclination ($\theta$) of the plane is $60^\circ$.
We need to find the coefficient of friction ($\mu$).
Analysis
Forces Acting on the Body:
When moving up the incline:
External force ($F_u$) acts up the incline.
Weight ($mg$) can be resolved into two components:
$mg \cos \theta$: Perpendicular to the plane.
$mg \sin \theta$: Parallel and down the plane.
Normal force ($N$): Perpendicular to the plane.
Frictional force ($f$) acts down the incline to oppose the motion.
When preventing the body from sliding down:
External force ($F_d$) acts up the incline.
Again, weight components:
$mg \cos \theta$: Perpendicular to the plane.
$mg \sin \theta$: Parallel and down the plane.
Normal force ($N$): Perpendicular to the plane.
Frictional force ($f$) now acts up the incline to oppose the motion.
Equating Forces:
Since there is no motion perpendicular to the incline, $N = mg \cos \theta$.
When moving up: $$ F_u = mg \sin \theta + f $$
The frictional force $f$ is: $ f = \mu N = \mu mg \cos \theta $. Hence, $$ F_u = mg \sin \theta + \mu mg \cos \theta $$
When preventing the body from sliding down: $$ F_d = mg \sin \theta - f $$ Similarly, $$ F_d = mg \sin \theta - \mu mg \cos \theta $$
Given Condition:
The force required to move up ($F_u$) is twice the force required to prevent sliding down ($F_d$): $$ F_u = 2 F_d $$
Substituting the expressions for $F_u$ and $F_d$: $$ mg \sin \theta + \mu mg \cos \theta = 2 (mg \sin \theta - \mu mg \cos \theta) $$
Solve for $\mu$:
Expand and simplify: $$ mg \sin \theta + \mu mg \cos \theta = 2 mg \sin \theta - 2 \mu mg \cos \theta $$ $$ mg \sin \theta + \mu mg \cos \theta = 2 mg \sin \theta - 2 \mu mg \cos \theta $$ $$ mg \sin \theta - 2 mg \sin \theta = -3 \mu mg \cos \theta $$ $$
mg \sin \theta = -3 \mu mg \cos \theta $$ Simplifying further: $$ \sin \theta = 3 \mu \cos \theta $$ $$ \mu = \frac{\sin \theta}{3 \cos \theta} = \frac{\tan \theta}{3} $$
Given $\theta = 60^\circ$:
$\tan 60^\circ = \sqrt{3}$ $$ \mu = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} $$
Conclusion
The coefficient of friction ($\mu$) is $\boxed{\frac{1}{\sqrt{3}}}$.
Thus, the correct option is C.
A projectile moving vertically upwards with a velocity of 200 m/s breaks into two equal parts at the height of 490 m. One part starts moving vertically upwards with a velocity of 400 m/s. How much time after the break up will the other part hit the ground?
A. √10 s
B. 2 √10s
C. 5 s
D. 10 s
To determine how much time after the breakup the other part of the projectile will take to hit the ground, we'll follow these steps carefully.
Step 1: Analyze the Given Information
Initial Scenario: A projectile moving vertically upwards with a velocity of 200 m/s.
Height at Breakup: The breakup occurs at a height of 490 m.
Breakup Details: The projectile splits into two equal parts. One part moves vertically upwards with a velocity of 400 m/s. We need to find the required time for the second part to hit the ground.
Step 2: Use Conservation of Momentum
When the projectile breaks into two equal parts, conservation of momentum must hold:
Total Momentum Before Breakup: ( m \times 200 )
After Breakup: If one part (mass ( m/2 )) moves with 400 m/s upward, the other part (mass ( m/2 )) will have a velocity ( v ).
Using the conservation of momentum: [ m \times 200 = \frac{m}{2} \times 400 + \frac{m}{2} \times v ]
Simplify this: [ 200m = 200m + \frac{mv}{2} ] [ 0 = \frac{mv}{2} ]
Thus, we get ( v = 0 ).
Conclusion: The second part does not have any initial velocity post-breakup and will start to fall freely.
Step 3: Calculate the Time to Fall
As the second part is in free fall from the height of ( 490 ) m with an initial velocity ( u = 0 ), we use the second equation of motion: [ h = ut + \frac{1}{2} g t^2 ] Here,
( h = -490 ) m (since the height is downward),
( u = 0 ),
( g = 9.8 , \text{m/s}^2 ).
Substitute the values: [ -490 = 0 + \frac{1}{2} \times 9.8 \times t^2 ] [ -490 = 4.9 t^2 ]
Solving for ( t ): [ t^2 = \frac{490}{4.9} = 100 ] [ t = \sqrt{100} = 10 , \text{seconds} ]
Conclusion
The time taken by the second part to hit the ground after the breakup is 10 seconds.
Answer:
Option (D) 10 s
A projectile of mass $M$ is fired so that the horizontal range is 4 km. At the highest point, the projectile explodes into two parts of masses $M / 4$ and $3M / 4$, respectively, and the heavier part starts falling down vertically with zero initial speed. The horizontal range (distance from the point of firing) of the lighter part is:
A 16 km B 1 km C 10 km D 2 km
To solve this problem, let's go step-by-step and use concepts related to projectile motion and center of mass.
Given
A projectile of mass $M$ is fired, reaching a horizontal range of 4 km.
At the highest point, it explodes into two parts: mass $\frac{M}{4} ) and ( \frac{3M}{4}$).
The heavier part falls vertically with zero initial speed.
Key Concept
In the absence of external forces, the horizontal displacement of the center of mass remains unchanged.
Breakdown
Initial Range: The total horizontal range, if there was no explosion, is 4 km.
Explosion at Highest Point:
The highest point is at half the total range due to the symmetry of projectile motion. So, at explosion, the center of mass is at$ \frac{4 \text{ km}}{2} = 2 \text{ km} $.
After Explosion
Let the horizontal range of the lighter part (mass $ \frac{M}{4} )) be ( x_1 $.
The heavier part (mass $ \frac{3M}{4} $) falls vertically, making its horizontal range still 2 km.
Center of Mass Calculations
To find the horizontal range of the lighter part $( x_1 )$, we will use the principle of conservation of momentum in the horizontal direction. Since no external forces act horizontally, the total horizontal displacement of the initial mass must equal the sum of the displacements of the exploded parts.
$$ M \times \text{initial center of mass displacement} = \frac{M}{4} \times x_1 + \frac{3M}{4} \times 2 \text{ km} $$
The initial center of mass displacement is the range of the projectile, which is 4 km:
$$ M \times 4 = \frac{M}{4} \times x_1 + \frac{3M}{4} \times 2 $$
Simplifying, we remove $ M $ from each term:
$$ 4 = \frac{1}{4} x_1 + \frac{3}{4} \times 2 $$
Simplifying further:
$$ 4 = \frac{1}{4} x_1 + 1.5 $$
Multiply through by 4 to clear the fraction:
$$ 16 = x_1 + 6 $$
Solving for $ x_1$:
$$ x_1 = 10 \text{ km} $$
Thus, the horizontal range of the lighter part is 10 km.
Final Answer
C: 10 km
Consider the following statements- Which of the statements given above is/are correct?
Think about the following statements:
The coldest place in India, Dras, is situated on the windward side of the Greater Himalayas.
Dras has been transformed into a significant tourist attraction Which one is/are correct among the above statements?
Option 1) Only 1
Option 2) Only 2
Option 3) Both 1 and 2
Option 4) Neither 1 nor 2
The correct option is B: 2 only
Explanation:
Statement 1 is incorrect.
The coldest place in India, Dras, is situated on the leeward side of the Greater Himalayas. On January 27, 2019, it recorded a seasonal low of -31.4 °C. However, the all-time record low for Dras is -43 °C.
If Dras were located on the windward side, the region would receive rainfall, which would prevent it from being the coldest place.
The climate in Dras is Mediterranean continental, influenced heavily by its altitude (3,300 meters). The average low temperature here is around -20 °C, and it can even reach -23 °C during the peak of winter.
Statement 2 is correct.
Dras has a population of only 1,201 individuals. The main ethnic groups in this area are the Shina and Balti people, with the majority following Islam.
The principal languages spoken in Dras are the Indo-Aryan languages, Shina and Balti.
Since the Kargil War in 1999, Dras has been transformed into a significant tourist attraction.
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Ask Chatterbot AINCERT Solutions - Dual nature of Radiation and Matter | NCERT | Physics | Class 12
Find the
(a) maximum frequency, and
(b) minimum wavelength of $\mathrm{X}$-rays produced by $30 \mathrm{kV}$ electrons.
Maximum Frequency $ f_{\text{max}} $: [ f_{\text{max}} = 7.253 \times 10^{18} \text{ Hz} ]
Minimum Wavelength $ \lambda_{\text{min}}$: [ \lambda_{\text{min}} = 4.136 \times 10^{-11} \text{ m} \quad \text{or} \quad 41.36 \text{ pm} ]
In summary, the maximum frequency of the X-rays produced by 30 kV electrons is $7.253 \times 10^{18} \text{ Hz}$, and the minimum wavelength is $41.36 \text{ pm}$.
The work function of caesium metal is $2.14 \mathrm{eV}$. When light of frequency $6 \times 10^{14} \mathrm{~Hz}$ is incident on the metal surface, photoemission of electrons occurs. What is the
(a) maximum kinetic energy of the emitted electrons,
(b) Stopping potential, and
(c) maximum speed of the emitted photoelectron?
Maximum Kinetic Energy of the Emitted Electrons$$ K_{\text{max}} = 0.547 \times 10^{-19} ; \text{J} $$
Stopping Potential$$ V_0 \approx 0.342 ; \text{V} $$
Maximum Speed of the Emitted Photoelectron$$ v_{\text{max}} \approx 3.47 \times 10^5 ; \text{m/s} $$
Therefore, the answers to the questions are:
(a) Maximum Kinetic Energy: $0.547 \times 10^{-19} ; \text{J}$
(b) Stopping Potential: $0.342 ; \text{V}$
(c) Maximum Speed of the Emitted Photoelectron: $3.47 \times 10^5 ; \text{m/s}$
The photoelectric cut-off voltage in a certain experiment is $1.5 \mathrm{~V}$. What is the maximum kinetic energy of photoelectrons emitted?
The maximum kinetic energy $ K_{\text{max}} $ of the photoelectrons emitted is: [ K_{\text{max}} = 2.40 \times 10^{-19} , \text{J} ]
So, **the maximum kinetic energy of the photoelectrons emitted is $ 2.40 \times 10^{-19}, \text{J}$ .
Monochromatic light of wavelength $632.8 \mathrm{~nm}$ is produced by a helium-neon laser. The power emitted is $9.42 \mathrm{~mW}$.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
(a) Energy and Momentum of Each Photon
Energy of Photon: [ E = 3.139 \times 10^{-19} , \text{J} ]
Momentum of Photon: [ p = 1.047 \times 10^{-27} , \text{kg} \cdot \text{m/s} ]
(b) Number of Photons Per Second
Since Wolfram|Alpha could not understand the initial complex query, let's break it down:
The energy of each photon: [ E = 3.139 \times 10^{-19} , \text{J} ]
The power of the laser beam: [ P = 9.42 , \text{mW} = 9.42 \times 10^{-3} , \text{J/s} ]
To find the number of photons $N$ emitted per second, we use: [ N = \frac{P}{E} ]
[ N = \frac{9.42 \times 10^{-3} , \text{J/s}}{3.139 \times 10^{-19} , \text{J}} = 3.00 \times 10^{16} , \text{photons/second} ]
(c) Speed of Hydrogen Atom matching Photon Momentum
Hydrogen Atom Momentum: $1.047 \times 10^{-27} , \text{kg} \cdot \text{m/s}$
The required velocity of a hydrogen atom to have this momentum: [ v = 12.55 , \text{cm/s} ]
Summary:
Energy of Photon: $3.139 \times 10^{-19} , \text{J}$
Momentum of Photon: $1.047 \times 10^{-27} , \text{kg} \cdot \text{m/s}$
Photons per Second: $3.00 \times 10^{16}$
Velocity of Hydrogen Atom for Matching Momentum: $12.55 , \text{cm/s}$
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{-15} \mathrm{~V}$ s. Calculate the value of Planck's constant.
The correct calculation is: [ h = (4.12 \times 10^{-15} , \text{Vs}) \times (1.602 \times 10^{-19} , \text{C}) ]
This gives: [ h = 6.60 \times 10^{-34} , \text{Js} ]
Therefore, the value of Planck's constant calculated from the given slope is approximately: [ h = 6.60 \times 10^{-34} , \text{Js} ]
The threshold frequency for a certain metal is $3.3 \times 10^{14} \mathrm{~Hz}$. If light of frequency $8.2 \times 10^{14} \mathrm{~Hz}$ is incident on the metal, predict the cutoff voltage for the photoelectric emission.
The cutoff voltage $ V_0 $ is:
$$ V_0 = 2.027 \text{ V} $$
Summary
For the given metal with a threshold frequency of $ 3.3 \times 10^{14} \text{ Hz} $, when light of frequency $ 8.2 \times 10^{14} \text{ Hz} $ is incident on it, the cutoff voltage for the photoelectric emission is 2.027 V.
The work function for a certain metal is $4.2 \mathrm{eV}$. Will this metal give hotoelectric emission for incident radiation of wavelength $330 \mathrm{~nm}$ ?
The energy of the incident photons is 3.762 eV.
The work function of the metal is 4.2 eV.
Since the energy of the incident photons $3.762 eV$ is less than the work function of the metal $4.2 eV$, the metal will not exhibit photoelectric emission for the given wavelength of 330 nm.
Light of frequency $7.21 \times 10^{14} \mathrm{~Hz}$ is incident on a metal surface. Electrons with a maximum speed of $6.0 \times 10^{5} \mathrm{~m} / \mathrm{s}$ are ejected from the surface. What is the threshold frequency for photoemission of electrons?
The threshold frequency $(v_0)$ for photoemission of electrons is:
[ \boxed{4.735 \times 10^{14} , \text{Hz}} ]
Light of wavelength $488 \mathrm{~nm}$ is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is $0.38 \mathrm{~V}$. Find the work function of the material from which the emitter is made.
Let's reinterpret the results correctly to get the work function $ \phi_0$:
The energy from the photons: [ h\nu = 4.07 \times 10^{-19} \mathrm{~J} ]
The energy contribution from the stopping potential: [ eV_0 = 6.09 \times 10^{-20} \mathrm{~J} ]
Now, we can find the work function: [ \phi_0 = h\nu - eV_0 ]
So, the calculation is: [ \phi_0 = 4.07 \times 10^{-19} \mathrm{~J} - 6.09 \times 10^{-20} \mathrm{~J} ]
Let's calculate this:
[ \phi_0 = 4.07 \times 10^{-19} \mathrm{~J} - 6.09 \times 10^{-20} \mathrm{~J} = 3.461 \times 10^{-19} \mathrm{~J} ]
The work function ( \phi_0 ) is approximately ( 3.461 \times 10^{-19} \mathrm{~J} ).
To express this in (\mathrm{eV}):
1 (\mathrm{eV} = 1.602 \times 10^{-19} \mathrm{~J}),
[ \phi_0 \text{ (in eV)} = \frac{3.461 \times 10^{-19} \mathrm{~J}}{1.602 \times 10^{-19} \mathrm{~J/eV}} \approx 2.16 \mathrm{~eV} ]
So, the work function of the material is 2.16 eV.
What is the de Broglie wavelength of
(a) a bullet of mass $0.040 \mathrm{~kg}$ travelling at the speed of $1.0 \mathrm{~km} / \mathrm{s}$,
(b) a ball of mass $0.060 \mathrm{~kg}$ moving at a speed of $1.0 \mathrm{~m} / \mathrm{s}$, and
(c) a dust particle of mass $1.0 \times 10^{-9} \mathrm{~kg}$ drifting with a speed of 2.2 $\mathrm{m} / \mathrm{s}$ ?
Here are the de Broglie wavelengths for the given objects:
(a) Bullet
Mass, $ m = 0.040 \ \text{kg} $
Speed, $ v = 1.0 \ \text{km/s} = 1000 \ \text{m/s} $
The de Broglie wavelength: $$ \lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34} \ \text{J} \cdot \text{s}}{0.040 \ \text{kg} \times 1000 \ \text{m/s}} \approx 1.66 \times 10^{-35} \ \text{m} $$
(b) Ball
Mass, $ m = 0.060 \ \text{kg}$
Speed, $ v = 1.0 \ \text{m/s} $
The de Broglie wavelength: $$ \lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34} \ \text{J} \cdot \text{s}}{0.060 \ \text{kg} \times 1.0 \ \text{m/s}} \approx 1.11 \times 10^{-32} \ \text{m} $$
(c) Dust Particle
Mass, $ m = 1.0 \times 10^{-9} \ \text{kg} $
Speed, $ v = 2.2 \ \text{m/s} $
The de Broglie wavelength: $$ \lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34} \ \text{J} \cdot \text{s}}{1.0 \times 10^{-9} \ \text{kg} \times 2.2 \ \text{m/s}} \approx 3.01 \times 10^{-25} \ \text{m} $$
Summary
Bullet:$\mathbf{1.66 \times 10^{-35} \ m} $
Ball:$\mathbf{1.11 \times 10^{-32} \ m} $
Dust Particle: $ \mathbf{3.01 \times 10^{-25} \ m}$
These wavelengths are extremely small, especially for macroscopic objects like a bullet and a ball. The dust particle, although still having a very small wavelength, shows a significantly larger de Broglie wavelength compared to the bullet and ball.
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
To demonstrate that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon), let's consider the following:
Electromagnetic Radiation Wavelength:The wavelength of electromagnetic radiation is typically denoted by $\lambda$.
Photon Energy and Momentum:According to the photon picture of light, the energy $E$ of a photon is given by: $$ E = h \nu $$ where ( h ) is Planck's constant and ( \nu ) is the frequency of the radiation.
The momentum ( p ) of the photon is given by: $$ p = \frac{h \nu}{c} $$ where ( c ) is the speed of light.
De Broglie Wavelength:According to de Broglie's hypothesis, the wavelength $\lambda$ associated with a particle of momentum ( p ) is: $$ \lambda = \frac{h}{p} $$
Let's calculate the de Broglie wavelength of a photon and see if it matches the wavelength of the electromagnetic radiation:
Substituting the photon's momentum into de Broglie's wavelength formula:
$$ \lambda_{\text{de Broglie}} = \frac{h}{p} = \frac{h}{\frac{h \nu}{c}} = \frac{h c}{h \nu} = \frac{c}{\nu} $$
The wavelength of electromagnetic radiation $ \lambda $ is related to the speed of light $ c $ and the frequency of the radiation $ \nu $ by:
$$ \lambda = \frac{c}{\nu} $$
Therefore, we see that:
$$ \lambda_{\text{de Broglie}} = \lambda $$
Conclusion:Both wavelengths are equal: $$ \lambda_{\text{de Broglie}} = \frac{c}{\nu} = \lambda $$
Hence, the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
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Ask Chatterbot AINotes - Dual nature of Radiation and Matter | Class 12 NCERT | Physics
Comprehensive Class 12 Notes on the Dual Nature of Radiation and Matter
The fascinating field of physics has revealed that radiation and matter have a dual nature, meaning they exhibit both wave-like and particle-like properties. Understanding this duality is essential for mastering the principles of modern physics. This comprehensive guide aims to explain the key concepts, experiments, and theories that shape our understanding of the dual nature of radiation and matter.
Introduction to Dual Nature of Radiation and Matter
The dual nature of radiation and matter refers to the fact that electromagnetic radiation (like light) and particles (like electrons) can exhibit both wave-like and particle-like properties. This duality is fundamental to quantum mechanics and helps explain various phenomena observed in the atomic and subatomic realms.
Maxwell’s Equations and Wave Nature of Light
James Clerk Maxwell’s equations established the wave nature of light. Hertz’s experiments further confirmed this by demonstrating the generation and detection of electromagnetic waves, laying the foundation for our modern understanding of light as an electromagnetic wave.
Historic Discoveries Leading to the Understanding of Atomic Structure
The late 19th century was a period of significant discoveries that advanced our knowledge of atomic structure.
Key Discoveries in Electric Discharge Experiments
Roentgen's discovery of X-rays and J. J. Thomson’s identification of the electron were pivotal. Thomson’s experiments with cathode rays led to the understanding that these rays were streams of fast-moving, negatively charged particles, known as electrons.
J. J. Thomson’s Experiment on Cathode Rays
Thomson determined the speed and specific charge (charge to mass ratio, e/m) of cathode ray particles, which travel at speeds ranging from 0.1 to 0.2 times the speed of light.
Photoelectric Effect
The photoelectric effect was discovered by Heinrich Hertz and further investigated by Wilhelm Hallwachs and Philipp Lenard. This phenomenon occurs when light irradiates a metal surface, causing electrons to be emitted.
Hertz’s Observations
Hertz observed the photoelectric effect during his experiments with electromagnetic waves. He found that ultraviolet light could enhance the discharge of electricity across a detector.
Further Investigations by Hallwachs and Lenard
Hallwachs and Lenard conducted detailed studies on the photoelectric effect. They discovered that current flowed in an evacuated glass tube when ultraviolet light hit the emitter plate, indicating the emission of electrons due to light.
Experimental Study of Photoelectric Effect
Effect of Intensity on Photocurrent
The photoelectric current is directly proportional to the intensity of incident light. As the intensity increases, more electrons are emitted per second.
Effect of Potential on Photocurrent
When the collector plate is positive relative to the emitter, the photoelectric current increases until it reaches a saturation point. If the collector plate is negative, the stopping potential is determined, beyond which no electrons reach the collector.
Effect of Frequency on Stopping Potential
The frequency of incident light affects the kinetic energy of photoelectrons. Higher frequencies result in higher stopping potentials, indicating that more energy is required to stop high-frequency photoelectrons.
Photoelectric Effect and Wave Theory of Light
Traditional wave theory could not explain several key aspects of the photoelectric effect, such as its immediate occurrence and the dependence on frequency rather than intensity.
Einstein’s Photoelectric Equation
Albert Einstein proposed that light consists of quanta (photons), each with energy hν (Planck's constant multiplied by frequency). Photoelectric emission occurs when an electron absorbs a photon's energy, exceeding the work function (ϕ₀) of the metal surface.
$K_{max} = hν - ϕ₀$
This simple equation elegantly explains all observations of the photoelectric effect.
Particle Nature of Light: The Photon
Einstein’s theory introduced the concept of the photon, with both energy and momentum, behaving like a particle. This particle-like behaviour was confirmed by subsequent experiments, including Compton scattering.
Wave Nature of Matter
Inspired by the dual nature of light, Louis de Broglie hypothesised that particles like electrons also exhibit wave-like properties. The de Broglie wavelength (λ) of a particle is given by:
$λ = h / p$
where h is Planck's constant and p is the momentum of the particle. This hypothesis was experimentally confirmed, revealing the wave-particle duality inherent in nature.
Summary of Dual Nature Concept
To summarise:
Wave Nature: Light exhibits wave-like properties, such as interference and diffraction.
Particle Nature: Light behaves as a stream of particles (photons) during interactions like the photoelectric effect.
Matter Waves: Particles like electrons exhibit wave-like properties under specific conditions.
Frequently Asked Questions
1. What is the dual nature of radiation and matter?The dual nature refers to the fact that radiation and matter exhibit both wave-like and particle-like properties.
2. What is the significance of J. J. Thomson's experiment?Thomson’s experiment provided the first determination of the speed and specific charge of electrons, leading to the identification of the electron as a fundamental particle.
3. How does Einstein’s photoelectric equation explain the photoelectric effect?Einstein proposed that light consists of quanta (photons), and photoelectric emission occurs when an electron absorbs enough photon energy to overcome the metal's work function.
4. What is the de Broglie hypothesis?De Broglie hypothesised that particles like electrons have an associated wavelength (λ = h/p), where h is Planck's constant and p is the momentum.
5. How do intensity and frequency affect photocurrent and stopping potential?The intensity of light affects the number of emitted electrons (photocurrent), while the frequency determines the kinetic energy of electrons and the stopping potential.
Understanding the dual nature of radiation and matter is essential for grasping the concepts of modern physics, paving the way for advancements in quantum mechanics and beyond.
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