Question

The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, ${\lambda }_1/{\lambda }_2$ of the photons emitted in this process is:

• A. $\frac{9}{7}$
• B. $\frac{7}{5}$
• C. $\frac{27}{5}$
• D. $\frac{20}{7}$

Answer 1

Answer: (D)

$\frac{20}{7}$

$\frac{1}{\lambda }=R(\frac{1}{n_f^2}-\frac{1}{n_i^2})$
here for wavelength${\lambda }_1,n_f=3,n_i=2$
$\frac{1}{{\lambda }_1}=R(\frac{1}{3^2}-\frac{1}{4^2})$
$\frac{1}{{\lambda }_1}=R\left(\frac{7}{9\times 16}\right)$
here for wavelength${\lambda }_1,n_f=2,n_i=3$
$\frac{1}{{\lambda }_2}=R(\frac{1}{2^2}-\frac{1}{3^2})$
$=R\left(\frac{5}{4\times 9}\right)$
$\frac{{\lambda }_1}{{\lambda }_2}=\frac{\frac{5}{36}}{\frac{7}{9\times 16}}$
Hence, from eqs. 1 and eqs. 2, we get
$=\frac{20}{7}$