Question

The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be

(Given ionization energy of $H$ = $2.18\times {10}^{-18}Jato{m}^{-1}$ and $h$ = $6.625\times {10}^{-34}Js$)

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

Answer 1

The correct option is A

$3.08\times {10}^{15}{s}^{-1}$

From Bohr's postulates, we know

[Frequency of radiation] $\nu$ = $3.289\times {10}^{15}\times Z^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]Hz$

Now, $\nu$ = $3.289\times {10}^{15}\times (1{)}^2[\frac{1}{1^2}-\frac{1}{4^2}]$

$\Rightarrow$ $\nu$ = $3.289\times {10}^{15}\times [1-\frac{1}{16}]$

$\Rightarrow$ $\nu$ = $3.289\times {10}^{15}\times \frac{15}{16}$

$\Rightarrow$ $\nu$ = $3.0834\times {10}^{15}Hz$