Question

If, in the first orbit of a hydrogen atom, the total energy of the electron is $-21.76\times {10}^{-19}J$, then the electric potential energy of an electron belonging to the second orbit of the same atom, will be,

• A. $-10.88\times {10}^{-19}\text{J}$
• B. $-5.44\times {10}^{-19}\text{J}$
• C. $-21.76\times {10}^{-19}\text{J}$
• D. $-43.52\times {10}^{-19}\text{J}$

Answer 1

The correct option is A $-10.88\times {10}^{-19}\text{J}$

The total energy of an electron, in hydrogen atom is,
$E_n=-\frac{m{e}^4}{8{ϵ}_0^2{h}^2{n}^2}$

$\Rightarrow E_n\propto \frac{1}{n^2}$

$\Rightarrow \frac{E_1}{E_2}=\frac{n_2^2}{n_1^2}$

$\frac{-21.76\times {10}^{-19}}{E_2}=\frac{4}{1}$

$\therefore E_2=-5.44\times {10}^{-19}\text{J}$

Now, potential energy and total energy are related as,
$|P.E.|=2|T.E.|$

$|P.E|=2|-5.44\times {10}^{-19}|=10.88\times {10}^{-19}\text{J}$

$\therefore P.E.=-10.88\times {10}^{-19}\text{J}$

Hence, $\left(A\right)$ is the correct answer.