Question

Which are correct for emission spectra of Balmer series in $H$-atom?

• A. ${\lambda }_{\left(innm\right)}=364.56\left[\frac{n_2^2}{n_2^2-n_1^2}\right]$; where $n_1=2$ and $n_2>2$
• B. $\frac{1}{\lambda }=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right];$ where $n_1=2$ and $n_2>2$; $R=3.29\times {10}^{15}{H}_z$
• C. $\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}];$ where $n_1=2$ and $n_2>2$; $R_H=1.09737\times {10}^{5}c{m}^{-1}$
• D. $\frac{1}{\lambda }=\frac{4c\left(inmse{c}^{-1}\right)}{364.56\times {10}^{-9}}[\frac{1}{n_1^2}-\frac{1}{n_2^2}];$ where $n_1=2$ and $n_2>2$;
  $c$ is speed of light.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\frac{1}{\lambda }=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right];$ where $n_1=2$ and $n_2>2$; $R=3.29\times {10}^{15}{H}_z$
B ${\lambda }_{\left(innm\right)}=364.56\left[\frac{n_2^2}{n_2^2-n_1^2}\right]$; where $n_1=2$ and $n_2>2$
C $\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}];$ where $n_1=2$ and $n_2>2$; $R_H=1.09737\times {10}^{5}c{m}^{-1}$
D $\frac{1}{\lambda }=\frac{4c\left(inmse{c}^{-1}\right)}{364.56\times {10}^{-9}}[\frac{1}{n_1^2}-\frac{1}{n_2^2}];$ where $n_1=2$ and $n_2>2$;

$c$ is speed of light.
From Bohr model, we know that
in a transition,
$\frac{1}{\lambda }=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$ where, $R_H=109700c{m}^{-}$
For Balmer series, electron gets deexcited from 3rd or upper level to second level so $n_1=2,n_2>2$.
Also we know that,
$E=hc/\lambda$ so
$v=\frac{4c\left(inmse{c}^{-1}\right)}{364.56\times {10}^{-9}}[\frac{1}{n_1^2}-\frac{1}{n_2^2}];$ where $n_1=2$ and $n_2>2$; $c$ is speed of light.