The angular momentum of an electron moving in an orbit of hydrogen atom is $1.5\left(\frac{\mathrm{h}}{\pi}\right)$. The energy in the same orbit is nearly

• A

  $-1.3 \mathrm{eV}$
• B

  $-1.4 \mathrm{eV}$
• Correct answer
  C

  $-1.5 \mathrm{eV}$
• D

  $-1.6 \mathrm{eV}$

Solution

Angular momentum of electron moving in $n^{\text {th }}$ orbit of hydrogen atom
Hence, $n=3$ Now, energy of electron in $3^{\text {rd }}$ orbit $\begin{aligned} E & =\frac{-13.6 \times z^2}{n^2} \\ & =\frac{-13.6 \times 1^2}{3^2} \\ & =-1.51 \mathrm{eV} \\ & \approx-1.5 \mathrm{eV}\end{aligned}$

Hence, the correct answer is option: 'C'.