The ratio of momentum of the photons of the 1 st and 2 nd line of Balmer series of Hydrogen atoms is α/β. The possible values of α and β are:- (a) 27 and 20 (b) 3 and 16 (c) 5 and 36 (d) 20 and 27

Answer: (d). Photon momentum p = (h)/(λ) ∝ (1)/(λ). 1st Balmer line ( 3→2): (1)/(λ₁) = R(1/4 - 1/9) = 5R/36 2nd Balmer line ( 4→2): (1)/(λ₂) = R(1/4 - 1/16) = 3R/16 (p₁)/(p₂) = (5/36)/(3/16) = 20/27 So the answer is (d) 20 and 27.

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Transitions, Photon Emission & Excitation

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

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Answer: (d)

Photon momentum $p = \frac{h}{\lambda} \propto \frac{1}{\lambda}$ .
1st Balmer line ( $3\to2$ ): $\frac{1}{\lambda_1} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}$
2nd Balmer line ( $4\to2$ ): $\frac{1}{\lambda_2} = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16}$
$\frac{p_1}{p_2} = \frac{5/36}{3/16} = \frac{20}{27}$
So the answer is (d) 20 and 27.

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