Question

Calculate the energy and frequency of the radiation emitted when an electron jumps from $n=3$ to $n=2$ in a hydrogen atom.

Answer 1

Calculating value of wavelength using Rydberg's Equation

For hydrogen like atoms, Rydberg’s equation is:

$\frac{1}{\lambda }=R_H(\frac{1}{n_i^2}-\frac{1}{n_f^2})$

Where, $\lambda =$ wavelength

$R_H=$ Rydberg’s constant $(1.09677\times {10}^7{m}^{-1})$

$n_f=3$ & $n_i=2$

Thus, $\frac{1}{\lambda }=1.09677\times {10}^7{m}^{-1}(\frac{1}{2^2}-\frac{1}{3^2})$

$\frac{1}{\lambda }=1.09677\times {10}^7{m}^{-1}\times \frac{5}{36}$

$\lambda =6.56\times {10}^{-7}m$

Calculating value of frequency and energy with the help of wavelength

Value of wavelength: $\lambda =6.56\times {10}^{-7}m$

Relation between wavelength and frequency is:

$v=\frac{c}{\lambda }$

$(c=3\times {10}^{8}m{s}^{-1},\lambda =6.56\times {10}^{-7}m)$

Thus,

$v=\frac{3\times {10}^{8}m/s}{6.56\times {10}^{-7}m}=0.46\times {10}^{15}{s}^{-1}$

$=4.57\times {10}^{16}Hz$

Relation between wavelength and energy is

$E=\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{6.56\times {10}^{-7}}$

$E=3.03\times {10}^{-19}J$

Conclusion:

Hence, the energy of radiation emitted is $3.03\times {10}^{-19}J$ and the frequency is $4.57\times {10}^{16}Hz$