Question

If the radius of electron orbit in the excited state of hydrogen atom is $476.1\, pm$, the energy of electron in that excited state in $J$ is
(Radius and energy of electron in the first orbit of hydrogen atom are $52.9\, pm$ and $-2.18 \times 10^{-18}\, J$ respectively)

• A. $- 2.42 \times 10^{-18}$
• B. $- 19.62 \times 10^{-18}$
• C. $- 2.42 \times 10^{-19}$
• D. $- 6.05 \times 10^{-19}$

Answer 1

Answer: (C)

Given : Radius of excited state of hydrogen atom $=476 . 1\, PM$

$E_{1} =2.18 \times 10^{-8} \, J/atom$

$E_{n} =-2.18 \times 10^{-18} \cdot \frac{Z^{2}}{n^{2}}$

$r_{1} =52.9\, pm$

$r_{n}=\frac{n^{2} \times 52.7}{Z} ; 476.1=\frac{n^{2} \times 52.7}{Z}$ [Z for hydrogen $=1$]

$\therefore n^{2}=9$ and $E_{n}=E_{3}=-\frac{2.18 Z^{2} \times 10^{-18} J}{9}$

$=-2.42 \times 10^{-19} J$