Question

The ratio of momentum of the photons of the $1^{\text{st}}$ and $2^{\text{nd}}$ line of Balmer series of Hydrogen atoms is $\alpha/\beta$. The possible values of $\alpha$ and $\beta$ are:-

• A. 27 and 20
• B. 3 and 16
• C. 5 and 36
• D. 20 and 27

Hint:

The momentum of a photon is proportional to the energy of the transition. Use the Balmer series formula with n=3 for the 1st line and n=4 for the 2nd line.

Answer 1

The Correct Option is D

According to Bohr's model of the Hydrogen atom, when an electron transitions between energy levels, it emits or absorbs a photon. The energy $E$ of the photon is given by:
$$ E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ eV} $$
The momentum $p$ of a photon is related to its energy $E$ by the relation $p = \frac{E}{c}$. Therefore, the momentum is directly proportional to the term $\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.

For the Balmer series, the final energy state is always $n_1 = 2$.

1. For the $1^{\text{st}}$ line of the Balmer series, the transition occurs from $n_2 = 3$ to $n_1 = 2$.
$$ p_1 \propto \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{9 - 4}{36} = \frac{5}{36} $$

2. For the $2^{\text{nd}}$ line of the Balmer series, the transition occurs from $n_2 = 4$ to $n_1 = 2$.
$$ p_2 \propto \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{4 - 1}{16} = \frac{3}{16} $$

3. Now, we find the ratio of these momenta:
$$ \frac{p_1}{p_2} = \frac{5/36}{3/16} = \frac{5}{36} \times \frac{16}{3} $$
Simplifying the fraction:
$$ \frac{p_1}{p_2} = \frac{5 \times 4}{9 \times 3} = \frac{20}{27} $$
Since the ratio is given as $\alpha/\beta$, we have $\alpha = 20$ and $\beta = 27$.