Question

Ratio between longest wavelength of $H$ atom in Lyman series to the shortest wavelength in Balmer series of $H{e}^{+}$ is

• A. $\frac{4}{3}$
• B. $\frac{36}{5}$
• C. $\frac{1}{4}$
• D. $\frac{5}{9}$

Answer 1

Answer: (A)

Longest wavelength in Lyman series of hydrogen atom arises from transition between $n=2$ and $n=1$ whose number is given by

${\bar{v}}_{H(2\to 1)}=R\times 1^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3}{4}R$

Shortest wavelength in Balmer series of $H{e}^{+}$ arises from transition between $n=\alpha$ and $n=2$ whose wave number is given by

${\bar{v}}_{He(\alpha -2)}=R\times 2^2\left(\frac{1}{2^2}-\frac{1}{{\alpha }^2}\right)$

$=R=\frac{{\bar{v}}_{He}}{{\bar{v}}_H}=\frac{{\lambda }_H}{{\lambda }_{He}}$

$\therefore \frac{{\lambda }_H}{{\lambda }_{He}}=\frac{4R}{3R}=\frac{4}{3}$