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

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.
$E=E_{4}-E_{1}$
$=\frac{- 13.6}{\left(\right. 4 \left.\right)^{2}}-\left[\frac{- 13.6}{\left(\right. 1 \left.\right)^{2}}\right]$
$=-0.85+13.6=12.75eV$
$\therefore $ The wavelength of the photon emitted is $\lambda =\frac{h c}{E}$ $\left[\because E = \frac{h c}{\lambda }\right]$
$=\frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{12.75 \times 1.6 \times 10^{- 19}}$
$=\frac{19.878 \times 10^{- 7}}{20.4}=0.974\times 10^{- 7}=974\overset{^\circ }{A}$