Question

If an electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom, then the frequency of emitted radiation in the hertz will be:
(Take Rydberg's constant, $R={10}^{5}c{m}^{-1}$)

• A. $\frac{3}{4}\times {10}^{15}$
• B. $\frac{3}{16}\times {10}^5$
• C. $\frac{3}{16}\times {10}^{15}$
• D. $\frac{9}{16}\times {10}^{15}$

Answer 1

The correct option is D $\frac{9}{16}\times {10}^{15}$

Using the relation,
$\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$={10}^{-5}(\frac{1}{2^2}-\frac{1}{4^2})={10}^{-5}\times (\frac{1}{4}-\frac{1}{16})$
$\lambda =\frac{16}{3}\times {10}^{-5}cm$
$\therefore$ Frequency, $\eta =\frac{c}{\lambda }=\frac{3\times {10}^{10}}{\frac{16}{3}\times {10}^{-5}}$
$=\frac{9}{16}\times {10}^{15}Hz$