Question

The ground state energy of hydrogen atom is $-13.6eV$ . If the electron jumps to the ground state from the 3^rd 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 $E_{\text{n}}=\frac{-13.6}{n^2}\text{eV}$
The required energy to jump electron to the ground state from the 3^rd excited state
$E=E_4-E_1$
$$\begin{gathered} =\frac{-13.6}{(4{)}^2}-[\frac{-13.6}{(1{)}^2}] \\ =-0.85+13.6=12.75\text{ eV} \end{gathered}$$
$\therefore$ The wavelength of the photon emitted is $\lambda =\frac{hc}{E}$ $[\because E=\frac{hc}{\lambda }]$
$$\begin{gathered} =\frac{6.626\times 10^{-34}\times 3\times 10^8}{12.75\times 1.6\times 10^{-19}} \\ =\frac{19.878\times 10^{-7}}{20.4}=0.974\times 10^{-7}=974\stackrel{{}^{\circ }}{A} \end{gathered}$$