Question

The wave number of last line of Lyman series of hydrogen spectrum is $109674c{m}^{-1}$. The wave number of $H_{\alpha }$ line in Balmer series of $H{e}^{+}$ is

• A. $438696c{m}^{-1}$
• B. $109674c{m}^{-1}$
• C. $30465c{m}^{-1}$
• D. $60930c{m}^{-1}$

Answer 1

Answer: (D)

$\bar{v}=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For the last line of Lyman series of $H$ -spectrum,

$Z=1,n_1=1$ and $n_2=\infty$

For $H_{\alpha }$ line in Balmer series of $H{e}^{+},Z=2,n_1=2$ and $n_2=3$

On putting the values,

$\bar{v}=109674\times 2^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=60930c{m}^{-1}$