Question

The frequency of radiation emitted when the electron falls from $n=4$ to $n=1$ in a hydrogen atom will be :
(Given, $h=6.625\times {10}^{-34}$Js):-

• A. $3.08\times {10}^{15}{s}^{-1}$
• B. $2.00\times {10}^{16}{s}^{-1}$
• C. $1.54\times {10}^{17}{s}^{-1}$
• D. $1.03\times {10}^{14}{s}^{-1}$

Answer 1

The correct option is A $3.08\times {10}^{15}{s}^{-1}$

$v=3.29\times {10}^{15}\{\frac{1}{n_1^2}-\frac{1}{n_1^2}\}se{c}^{-1}$
$v=3.29\times {10}^{15}\{\frac{1}{1^2}-\frac{1}{4^2}\}se{c}^{-1}$
$v=3.29\times {10}^{15}\times \frac{15}{16}se{c}^{-1}=3.08\times {10}^{15}se{c}^{-1}$