Atomic Structure - Result Question 105
####81. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition $n=4$ to $n=2$ of $He^{+}$spectrum?
(1993, 3M)
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Solution:
- The expression for transition wavelength is given by Rydberg’s equation :
$$ \frac{1}{\lambda}=R _H Z^{2}\left(\frac{1}{n _1^{2}}-\frac{1}{n _2^{2}}\right) $$
Equating the transition wavelengths of $H-$ atom and $He^{+}$ion,
$$ R _H\left(\frac{1}{n _1^{2}}-\frac{1}{n _2^{2}}\right)=R _H\left(\frac{4}{2^{2}}-\frac{4}{4^{2}}\right) $$
Equating termwise on left to right of the above equation gives $n _1=1$ and $n _2=2$
