Question

An electron jumps from the 4^th orbit to the 2^nd orbit of hydrogen atom. Given : the Rydberg’s constant R = 10⁵ cm^–1. The frequency in Hz of the emitted radiation will be –

• A. $\frac{3}{16}$× 10⁵
• B. $\frac{3}{6}$× 10¹⁵
• C. $\frac{9}{16}$× 10¹⁵
• D. $\frac{3}{4}$× 10¹⁶

Answer 1

Answer: (C)

$\frac{9}{16}$× 10¹⁵

Answer 2

$\frac{1}{\lambda}$ = R$\left\lbrack \frac{1}{2^{2}}–\frac{1}{4^{2}} \right\rbrack$ or $\frac{\nu}{c}$ = R$\left\lbrack \frac{1}{4}–\frac{1}{16} \right\rbrack$

or n = cR$\left\lbrack \frac{1}{4}–\frac{1}{16} \right\rbrack$

3 × 10⁸×10⁵(10^–2)^–1×$\frac{3}{16}$ = $\frac{9}{16}$× 10¹⁵ Hz