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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 wavelengths $\lambda_1:\lambda_2$ emitted in the two cases is

Option: 1
$7/5$

Option: 2
$27/20$

Option: 3
$27/5$

Option: 4
$20/7$

Answers (1)

best_answer

Here, for wavelength $\lambda_1,$

$\mathrm{n_1=4 \text { and } n_2=3 \text {. }}$

And for $\mathrm{\lambda_2, n_1=3 \: and\: n_2=2.}$

We have, $\mathrm{\frac{h c}{\lambda}=-13.6\left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right]}$

So, for $\mathrm{\lambda_1}$

$\mathrm{\Rightarrow \quad \frac{h c}{\lambda_1} =-13.6\left[\frac{1}{(4)^2}-\frac{1}{(3)^2}\right] }$

$\mathrm{\frac{h c}{\lambda_1} =13.6\left[\frac{7}{144}\right]}$ ......(1)

Similarly, for $\mathrm{\lambda_2}$

$\mathrm{\Rightarrow \quad \frac{h c}{\lambda_c} =-13.6\left[\frac{1}{(3)^2}-\frac{1}{(2)^2}\right] }$

$\mathrm{\frac{h c}{\lambda_2} =13.6\left[\frac{5}{36}\right]}$

Hence, from Eqs. (i) and (ii), we get

$\mathrm{ \frac{\lambda_1}{\lambda_2}=\frac{20}{7} }$

Hence option 4 is correct.

Posted by

Kuldeep Maurya

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