Atomic Structure - Result Question 102
####78. Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen(1996, 1M)
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Answer:
Correct Answer: 78. $\left(3.66 \times 10^{-5} cm\right)$ 87. $(1220 \AA)$
Solution:
- The Rydberg’s equation for $H$-atom is
$$ \frac{1}{\lambda}=\bar{v}(\text { wave number })=R _H\left(\frac{1}{n _1^{2}}-\frac{1}{n _2^{2}}\right) $$
For Balmer series, $n _1=2$ and $n _2=3,4,5, \ldots, \infty$
For shortest $\lambda, n _2$ has to be maximum, i.e. infinity. Then
$$ \bar{v}=R _H\left(\frac{1}{4}-\frac{1}{\infty}\right)=\frac{R _H}{4}=\frac{1.09 \times 10^{7}}{4}=2.725 \times 10^{6} m^{-1} $$
