Question

The ground state energy of the hydrogen atom is $-13.6eV$ . If the electron jumps to the ground state from the third excited state, the wavelength of the emitted photon is

• A. $875\stackrel{{}^{\circ }}{A}$
• B. $1052\stackrel{{}^{\circ }}{A}$
• C. $752\stackrel{{}^{\circ }}{A}$
• D. $974\stackrel{{}^{\circ }}{A}$

Answer 1

Answer: (D)

The energy of an electron in $n^{th}$ orbit is given by,
$E_n=\frac{-13.6}{n^2}eV$ .
The required energy to jump electron to the ground state from the third excited state.
$E=E_4-E_1$
$=\frac{-13.6}{(4{)}^2}-\left[\frac{-13.6}{(1{)}^2}\right]$
$=-0.85+13.6=12.75eV$
$\therefore$ The wavelength of the photon emitted is $\lambda =\frac{hc}{E}$ $\left[\because E=\frac{hc}{\lambda }\right]$
$=\frac{6.626\times 10^{-34}\times 3\times 10^8}{12.75\times 1.6\times 10^{-19}}$
$=\frac{19.878\times 10^{-7}}{20.4}=0.974\times 10^{-7}=974\stackrel{{}^{\circ }}{A}$