The shortest wavelength of Hatom in the Lyman series is

Question:

The shortest wavelength of Hatom in the Lyman series is $\lambda_{1}$. The longest wavelength in the Balmer series is $\mathrm{He}^{+}$is :

  1. $\frac{36 \lambda_{1}}{5}$

  2. $\frac{5 \lambda_{1}}{9}$

  3. $\frac{9 \lambda_{1}}{5}$

  4. $\frac{27 \lambda_{1}}{5}$


Correct Option:

Solution:

Shortest wavelength $\rightarrow$ Max. energy $(\infty \rightarrow 1)$

For Lyman series of H atom,

$\frac{1}{\lambda_{1}}=R_{\mathrm{H}}(1)^{2}\left[\frac{1}{1}-0\right]$

$\Rightarrow \frac{1}{\lambda_{1}}=R_{\mathrm{H}} \Rightarrow R_{\mathrm{H}}=\frac{1}{\lambda_{1}}$

For Balmer series of $\mathrm{He}^{+}$,

$\frac{1}{\lambda}=R_{\mathrm{H}}(2)^{2}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right] \Rightarrow \frac{1}{\lambda}=R_{\mathrm{H}}(4)\left(\frac{9-4}{36}\right)$

$\Rightarrow \frac{1}{\lambda}=\frac{5 R_{\mathrm{H}}}{9} \Rightarrow \lambda=\frac{9}{5 R_{\mathrm{H}}}=\frac{9 \lambda_{1}}{5}$

 

 

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