Question

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

Answer 1

The $n_i$ = 4 to $n_f$ = 2 transition will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,
$E=2.18\times {10}^{-18}[\frac{1}{n_i^2}-\frac{1}{n_f^2}]$
Substituting the values in the given expression of E:
$E=2.18\times {10}^{-18}[\frac{1}{4^2}-\frac{1}{2^2}]=2.18\times {10}^{-18}\left[\frac{1-4}{16}\right]=2.18\times {10}^{-18}\times (-\frac{3}{16})$
E = – (4.0875 × ${10}^{–19}$ J)
The negative sign indicates the energy of emission.
Wavelength of light emitted $\left(\lambda \right)=\frac{hc}{E}(sinceE=\frac{hc}{\lambda })$
Substituting the values in the given expression of λ:
$\lambda =\frac{(6.626\times {10}^{-34})(3\times {10}^8)}{4.0875\times {10}^{-19}}\lambda =4.8631\times {10}^{-7}m\lambda =4.8631\times {10}^{-9}m\lambda =486nm$