Question

The wavenumber of last line of Lyman series of hydrogen spectrum is $109674{\text{cm}}^{-1}$. The wave number of $H-\alpha$ line in Balmer series $H{e}^{+}$ is :

• A. $438696{\text{cm}}^{-1}$
• B. $106974{\text{cm}}^{-1}$
• C. $30465{\text{cm}}^{-1}$
• D. $60930{\text{cm}}^{-1}$

Answer 1

The correct option is D $60930{\text{cm}}^{-1}$

For last line of Lyman series of $H-$ Spectrum;
$Z=1,n_1=1,n_2=\infty$
We know,
$\overline{v}=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
So, according to the question
$109674=R\times 1[\frac{1}{1^2}-\frac{1}{{\infty }^2}]$
$\Rightarrow R=109674c{m}^{-1}$

For $H\alpha$ line in Balmer series of $H{e}^{+};Z=2,n_1=2,n_2=3$
Substitting the values we get,
$\overline{v}=109674\times 2^2[\frac{1}{2^2}-\frac{1}{3^2}]$
$\Rightarrow \overline{v}=109674\times 2^2\left[\frac{5}{36}\right]$
On solving, we get
$\overline{v}=60930{\text{cm}}^{-1}$