Atoms Ques 12
12. Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelength $\lambda_1: \lambda_2$ emitted in the two cases is
[2012]
(a) $7 / 5$
(b) $27 / 20$
(c) $27 / 5$
(d) $20 / 7$
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Answer:
Correct Answer: 12.(d)
Solution:
- (d)
According to Rydberg formula
$\frac{1}{\lambda}=R[\frac{1}{n_f^{2}}-\frac{1}{n_i^{2}}]$
In first case, $n_f=3, n_i=4$
$\therefore \frac{1}{\lambda_1}=R[\frac{1}{3^{2}}-\frac{1}{4^{2}}]=R[\frac{1}{9}-\frac{1}{16}]=\frac{7}{144} R \quad$ ……(i)
In second case, $n_f=2, n_i=3$
$\therefore \frac{1}{\lambda_2}=R[\frac{1}{2^{2}}-\frac{1}{3^{2}}]=R[\frac{1}{4}-\frac{1}{9}]=\frac{5}{36} R \quad$ …..(ii)
Divide (ii) by (i), we get
$\frac{\lambda_1}{\lambda_2}=\frac{5}{36} \times \frac{144}{7}=\frac{20}{7}$
BETA
