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

The momentum $p$ of a photon is given by de Broglie's relation $p = \frac{h}{\lambda}$. In the spectrum of Hydrogen, the reciprocal of the wavelength $\frac{1}{\lambda}$ is given by the Rydberg formula:
$$ \frac{1}{\lambda} = R \left( \frac{1}{n_l^2} - \frac{1}{n_h^2} \right) $$
where $R$ is the Rydberg constant. Thus, $p = hR \left( \frac{1}{n_l^2} - \frac{1}{n_h^2} \right)$, meaning momentum is proportional to $\left( \frac{1}{n_l^2} - \frac{1}{n_h^2} \right)$.

Balmer series transitions end at $n_l = 2$.
- The first line corresponds to the transition from $n = 3 \to n = 2$.
- The second line corresponds to the transition from $n = 4 \to n = 2$.

Calculate the momentum factor for the first line ($p_1$):
$$ p_1 \propto \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = \frac{1}{4} - \frac{1}{9} = \frac{5}{36} $$

Calculate the momentum factor for the second line ($p_2$):
$$ p_2 \propto \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = \frac{1}{4} - \frac{1}{16} = \frac{3}{16} $$

Calculate the ratio $p_1 : p_2$:
$$ \frac{p_1}{p_2} = \frac{5/36}{3/16} = \frac{5 \times 16}{36 \times 3} = \frac{80}{108} $$
Reducing to simplest form by dividing by 4:
$$ \frac{p_1}{p_2} = \frac{20}{27} $$
Comparing with $\alpha/\beta$, we find $\alpha = 20$ and $\beta = 27$.