# Dual nature of Radiation and Matter - Class 12 - Physics

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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

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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