Question

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

• A. $438696 \,cm ^{-1}$
• B. $109674 \,cm ^{-1}$
• C. $30465\, cm ^{-1}$
• D. $60930\, cm ^{-1}$

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 $He ^{+}, Z=2, n_{1}=2$ and $n_{2}=3$

On putting the values,

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