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. $\frac{9}{5}$

Answer 1

Answer: (D)

$\frac{9}{5}$

Wave number $=R[{\left(\frac{1}{n_1}\right)}^2-{\left(\frac{1}{n_2}\right)}^2]=\frac{1}{wavelength}$

So, $\frac{1}{\lambda }={RZ}^2(\frac{1}{2^2}-\frac{1}{3^2})$

$=R\times 4\times \left(\frac{5}{36}\right)$

$\frac{1}{\lambda }$ $=\frac{5}{9}R$

So, the ratio is$\lambda$= $\frac{9}{5}R$.