Question

What is the ratio of energy of an $e^{-}$, in the $2^{\text{nd}}$ orbit of hydrogen ion to that of electron in the $4^{\text{th}}$ orbit of helium ion ?

• A. $2$
• B. $4$
• C. $16$
• D. $1$

Answer 1

The correct option is D $1$

Total energy of an electron is,
$E=-K\left(\frac{z^2}{n^2}\right)$
energy of $e^{-}$ in $2^{\text{nd}}$ orbit of H atom is, $E_1=-k\left(\frac{1^2}{2^2}\right)=-K\left(\frac{1}{4}\right)$

Similarly, energy of $e^{-}$ in the $4^{\text{th}}$ orbit of the He atom is,
$E_2=-K\left(\frac{2^2}{4^2}\right)=\frac{-K}{4};$

So, they both are equal.

$\therefore$ The ratio will be $1$.

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

Key concept : The energy (PE + KE) of $e^{-}$ in a $H^{+}$ atom in the first orbit will be $-13.6\text{eV}$ $\therefore$ The energy for any orbit will be $-13.6\left(\frac{z^2}{n^2}\right)\text{eV}$