Question

Calculate the wavelength and energy of radiation emitted for the electronic transition from infinite to stationary state of hydrogen atom. (Given: $R=1.09678\times {10}^7{m}^{-1},h=6.6256\times {10}^{-34}$ J-s\ and $c=2.9979\times {10}^8{m}^{-1}$)

Answer 1

$h_2=\infty n_1=1$

$\frac{1}{\lambda }=RH(\frac{1}{{h_1}^2}-\frac{1}{{h_2}^2})$

$\frac{1}{\lambda }=RH(\frac{1}{1^2}-\frac{1}{{\infty }^2})$

$\lambda =\frac{1}{R_H}=\frac{1}{1.9678\times {10}^7}m=916nm$

$△E=-13.6(\frac{1}{1^2}-\frac{1}{{\infty }^2})eV$

$=-13.6eV$