A.

2:1

B.

4:1

C.

1:2

D.

1:4

Answers (1)

Here, the two particales have the same mass m, so the reducted mass is \mu=\frac{m m}{m+m}=\frac{m^{2}}{2 m}=\frac{m}{2}$

where m is the electron mass.

\text { We know that } E_{n} \propto m \Rightarrow \frac{E_{n}^{\prime}}{E_{n}}=\frac{\mu}{m}=\frac{1}{2} 

Also E = \frac{hc}{\lambda }  , wavelength is inversely proportional to Energy. Therefore \frac{\lambda' _n}{\lambda_n} = \frac{2}{1} .

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