atoms Question 15
Question: Q. 6. Calculate the shortest wavelength of the spectral lines emitted in Balmer series. (Given Rydberg constant, $R=10^{7} \mathrm{~m}^{-1}$ )
R] [Delhi OD 2016]
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Solution:
Ans.
$$ \frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right) $$
For shortest wavelength, $n=\infty$
Therefore, $\frac{1}{\lambda}=\frac{R}{4}$
$$ \Rightarrow \quad \lambda=\frac{4}{R}=4 \times 10^{-7} \mathrm{~m} $$
[CBSE Marking Scheme 2016]
