The angular momentum of an electron moving in an orbit of hydrogen atom is $1…

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
  1. $-1.3 \mathrm{eV}$
  2. $-1.4 \mathrm{eV}$
  3. $-1.5 \mathrm{eV}$
  4. $-1.6 \mathrm{eV}$

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 $\begin{aligned} E & =\frac{-13.6 \times z^2}{n^2} \\ & =\frac{-13.6 \times 1^2}{3^2} \\ & =-1.51 \mathrm{eV} \\ & \approx-1.5 \mathrm{eV}\end{aligned}$

Asked in: NEET 2023 (Manipur)

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