MCQ
The frequency corresponding to transition $n = 2$ to $n = 1$ in hydrogen atom is
- A$15.66 \times {10^{10}}Hz$
- ✓$24.66 \times {10^{14}}Hz$
- C$30.57 \times {10^{14}}Hz$
- D$40.57 \times {10^{24}}Hz$
✓
Answer
Correct option: B.
$24.66 \times {10^{14}}Hz$
b
(b) $v = \frac{1}{\lambda } = R\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right] = 109678\left[ {\frac{1}{1} - \frac{1}{4}} \right] = 82258.5$
(b) $v = \frac{1}{\lambda } = R\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right] = 109678\left[ {\frac{1}{1} - \frac{1}{4}} \right] = 82258.5$
$\lambda = 1.21567 \times {10^{ - 5}}cm\;\;\;{\rm{or}}\;\;\;\lambda = 12.1567 \times {10^{ - 6}}cm$
$ = 12.1567 \times {10^{ - 8}}\,m$
$v = \frac{c}{\lambda } = \frac{{3 \times {{10}^8}}}{{12.567 \times {{10}^{ - 8}}}} = 24.66 \times {10^{14}}Hz$.
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