Question

If the radius of electron orbit in the excited state of hydrogen atom is $476.1pm$, the energy of electron in that excited state in $J$ is
(Radius and energy of electron in the first orbit of hydrogen atom are $52.9pm$ 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 $=476.1PM$

$E_1=2.18\times 10^{-8}J/atom$

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

$r_1=52.9pm$

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

$\therefore n^2=9$ and $E_n=E_3=-\frac{2.18Z^2\times 10^{-18}J}{9}$

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