Question

The ground state energy of hydrogen atom is $-13.6 \, eV$ . If the electron jumps to the ground state from the 3^rd 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 $E_{\text{n}} = \frac{- 13.6}{n^{2}} \, \, \text{eV}$
The required energy to jump electron to the ground state from the 3^rd excited state
$E = E_{4} - E_{1}$
$=\frac{- 13.6}{\left(\right. 4 \left.\right)^{2}}-\left[\right. \frac{- 13.6}{\left(\right. 1 \left.\right)^{2}} \left]\right. \\ =-0.85+13.6=12.75\text{ eV}$
$\therefore $ The wavelength of the photon emitted is $\lambda = \frac{h c}{E}$ $\left[\right. \because E = \frac{h c}{\lambda } \left]\right.$
$=\frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{12.75 \times 1.6 \times 10^{- 19}} \\ =\frac{19.878 \times 10^{- 7}}{20.4}=0.974\times 10^{- 7}=974\overset{^\circ }{\text{A}}$