Question

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

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

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

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

$=R=\frac{\bar{v}_{ He }}{\bar{v}_{ H }}=\frac{\lambda_{ H }}{\lambda_{ He }}$

$\therefore \frac{\lambda_{ H }}{\lambda_{ He }}=\frac{4 R}{3 R}=\frac{4}{3}$