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(A) The energy of a photon is given by $E = mc^2 = \frac{hc}{\lambda}$,which implies $m = \frac{h}{\lambda c}$. Thus,$m \propto \frac{1}{\lambda}$.For

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$27 : 5$

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(A) The energy of a photon is given by $E = mc^2 = \frac{hc}{\lambda}$,which implies $m = \frac{h}{\lambda c}$. Thus,$m \propto \frac{1}{\lambda}$.For the Lyman series,the first line corresponds to the transition from $n_2 = 2$ to $n_1 = 1$:$\frac{1}{\lambda_1} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R_H \left( 1 - \frac{1}{4} ight) = \frac{3R_H}{4}$.For the Balmer series,the first line corresponds to the transition from $n_2 = 3$ to $n_1 = 2$:$\frac{1}{\lambda_2} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R_H \left( \frac{1}{4} - \frac{1}{9} ight) = \frac{5R_H}{36}$.Since $m \propto \frac{1}{\lambda}$,the ratio of masses is $\frac{m_1}{m_2} = \frac{\lambda_2}{\lambda_1} = \frac{1/\lambda_1}{1/\lambda_2} = \frac{3R_H/4}{5R_H/36} = \frac{3}{4} 	imes \frac{36}{5} = \frac{27}{5}$.Therefore,the ratio is $27 : 5$.

The masses of photons corresponding to the first lines of Lyman and Balmer series of the spectrum of hydrogen atom are in the ratio of

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