The energy of an electron in first Bohr orbit of $\mathrm{H}-$ atom is $-13.6 \mathrm{eV}$. The energy value…
- $-27.2 \mathrm{eV}$
- $30.6 \mathrm{eV}$
- $-30.6 \mathrm{eV}$
- $27.2 \mathrm{eV}$
Solution
$\begin{aligned} E &=-13.6 \times \frac{Z^{2}}{n^{2}}=-13.6 \times \frac{(3)^{2}}{(2)^{2}} \\ &=\frac{-13.6 \times 9}{4}=-30.6 \mathrm{eV} \end{aligned}$
Asked in: JEE-TOPICTESTS-CHEMISTRY