Question

The frequency of radiation emitted when the electron falls from $n=4$ to $n=1$ in a hydrogen atom will be (Given : ionisation energy of $H=2.18\times {10}^{-18}J{atom}^{-1}$ and $h=6.626\times {10}^{-34}J-s$)

• 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. $2\times {10}^{15}{s}^{-1}$

Answer 1

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

The frequency of radiation emitted when the electron falls from $n=4$ to $n=1$ in a hydrogen atom will be $3.08\times {10}^{15}{s}^{-1}$
$I.E=-E_1=2.18\times {10}^{-18}J/atom$
$E_1=-I.E=-2.18\times {10}^{-18}J/atom$
$E_n=\frac{E_1}{n^2}$
$E_4=\frac{E_1}{4^2}$
$E_4=\frac{-2.18\times {10}^{-18}J/atom}{4^2}$
$E_4=-1.36\times {10}^{-19}J/atom$
$\Delta E=E_1-E_4$
$\Delta E=-2.18\times {10}^{-18}J/atom-(-1.36\times {10}^{-19}J/atom)$
$\Delta E=-2.04\times {10}^{-18}J/atom$
The frequency $\nu =\frac{\Delta E}{h}$
$\nu =\frac{2.04\times {10}^{-18}J/atom}{6.626\times {10}^{-34}Js}$
$\nu =3.08\times {10}^{15}{s}^{-1}$