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(D) The Balmer series corresponds to $n_f = 2$. The 1st line corresponds to $n_i = 3$,and the 2nd line corresponds to $n_i = 4$. Momentum $p = E/c = (

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$20$ and $27$

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(D) The Balmer series corresponds to $n_f = 2$. The 1st line corresponds to $n_i = 3$,and the 2nd line corresponds to $n_i = 4$. Momentum $p = E/c = (h
u)/c = h/\lambda$. Since $1/\lambda = R(1/2^2 - 1/n_i^2)$,we have $p \propto (1/4 - 1/n_i^2)$. For the 1st line,$p_1 \propto (1/4 - 1/9) = 5/36$. For the 2nd line,$p_2 \propto (1/4 - 1/16) = 3/16$. The ratio $p_1/p_2 = (5/36) / (3/16) = (5/36) 	imes (16/3) = (5 	imes 4) / (9 	imes 3) = 20/27$. Thus,$\alpha = 20$ and $\beta = 27$.

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

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