# Atoms - Class 12 - Physics

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## Extra Questions - Atoms | NCERT | Physics | Class 12

A stone is dropped from a height of 125 m. After 3 sec of its fall, it is stopped for a moment, and then released. The total time of flight is:

A) 5 sec

B) 6 sec

C) 9 sec

D) 7 sec

To determine the total flight time of a stone dropped from a height of 125 m and momentarily stopped after 3 seconds, we follow these steps:

**Calculate the distance fallen in the first 3 seconds**:

The distance $S$ covered in free fall is given by the equation: $$ S = \frac{1}{2} g t^2 $$ Here, $g = 10 , \text{m/s}^2$ (assuming a rounded value for gravity) and $t = 3 , \text{s}$. Thus, $$ S = \frac{1}{2} \times 10 \times (3^2) = 45 , \text{m} $$ This means after 3 seconds, the stone has fallen 45 m.**Determine the remaining distance**: The initial height from which the stone was dropped was $125 , \text{m}$. After falling 45 m, the remaining distance to the ground is: $$ 125 , \text{m} - 45 , \text{m} = 80 , \text{m} $$**Calculate the time to fall the remaining distance**: Using the same formula for free fall: $$ S = \frac{1}{2} g t^2 $$ We substitute $S = 80 , \text{m}$: $$ 80 = \frac{1}{2} \times 10 \times t^2 $$ Solving for $t^2$: $$ t^2 = \frac{160}{10} = 16 , \text{s}^2 \Rightarrow t = 4 , \text{s} $$ Therefore, it takes an additional 4 seconds to fall the remaining 80 m.**Compute the total time of flight**: The stone falls for 3 seconds, is stopped at that point, and resumes to fall for another 4 seconds, giving a**total flight time of 7 seconds**.

Based on this analysis, the answer is:

**D) 7 sec**

What happens in a $\beta^+$ (beta positive) decay?

A) Proton converts into a neutron.

B) Neutron converts into a proton.

C) Electron and antineutrino are created.

D) Positron and neutrino are created.

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In the atom of an element, seven electrons are present in the outermost shell. If it acquires an inert gas configuration by receiving more electrons, the charge on the ion formed will be:

(A) $7-$

(B) $1+$

(C) $7+$

(D) $1-$.

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Which of the following is (are) correct for a $\mathrm{H}_{2}^{\oplus}$-like species?

A. The energy gap between the consecutive energy orbits decreases as the value of "$n$" increases.

B. The longest wavelength in any spectral series corresponds to the $\alpha$-line in that series.

C. Each spectral series is bounded by minimum and maximum wavelengths, and the range follows a continuous distribution as given by Bohr's theory.

D. Kinetic energy of the electron decreases, whereas the potential energy increases as the value of "$n$" increases.

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The maximum number of electrons that can be filled in an orbit (shell) can be found by using the formula $2n^{2}$, where '$n$' stands for an orbit's serial number. Calculate the maximum number of electrons that can be accommodated in the $2^{\text{nd}}$ orbit (shell) of an atom.

A. 4 electrons

B. 6 electrons

C. 8 electrons

D. 12 electrons

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"The ratio of the difference between the $3^{\text{rd}}$ and $2^{\text{nd}}$ orbit of $\mathrm{H}$ atom and $\mathrm{Li}^{2+}$ ion is:

A) $5:15$

B) $15:5$

C) $5:3$

D) $3:5"