Question

The frequency of radiation emitted when the electron falls from $n=4$ to $n=$ lina hydrogen atom will be (Given ionisation energy of $H=2.18{10}^{-18}Jato{m}^{-1}$ and $h=6.625\times {10}^{-34}Js)$

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

Answer 1

Answer: (C)

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

Solution:- (C) $3.08\times {10}^{15}{s}^{-1}$

As we know that,

$\Delta E=(2.18\times {10}^{-18})\times [\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}]$

As the electron falls from n=4 to n= 1.

$\therefore n_1=1;n_2=4$

$\therefore \Delta E=2.18\times {10}^{-18}\times [\frac{1}{1^2}-\frac{1}{4^2}]$

$\Rightarrow \Delta E=2.18\times {10}^{-18}\times \frac{15}{16}=2.043$

As we know that frequency and energy can be related as;

$\Delta E=h\nu$

$\Rightarrow \nu =\frac{\Delta E}{h}=\frac{2.043\times {10}^{-18}}{6.625\times {10}^{-34}}=0.308\times {10}^{16}=3.08\times {10}^{15}{s}^{-1}$