Atomic Structure Result Question 81-1
81. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition $n=4$ to $n=2$ of $\mathrm{He}^{+}$ spectrum?
(1993, 3M)
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Answer:
Correct Answer: 81. ( a )
Solution:
- The expression for transition wavelength is given by Rydberg’s equation :
$ \dfrac{1}{\lambda}=R_{\mathrm{H}} Z^2\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right) $
Equating the transition wavelengths of $H $-atom and $\mathrm{He}^{+}$ ion,
$ R_{\mathrm{H}}\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)=R_{\mathrm{H}}\left(\dfrac{4}{2^2}-\dfrac{4}{4^2}\right) $
Equating termwise on left to right of the above equation gives $n_1=1$ and $n_2=2$
BETA
