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

Answer 1

Answer: (B)

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

$E$= Energy involved= $E_o\times [\frac{1}{(1{)}^2}-\frac{1}{(4{)}^2}]$

$=E_o\times \frac{15}{16}$

where, $E_o=$ Ionization energy

$\because E=h\nu$

$\therefore h\nu =\frac{15E_o}{16}$

$\therefore \nu$=frequency= $\frac{15E_o}{16h}=\frac{15\times 2.18\times {10}^{-18}}{16\times 6.625\times {10}^{-34}}=3.08\times {10}^{15}{s}^{-1}$