The energy of an electron in an orbit of the Bohr s ato... | null (Physics) | DivineJEE

JEE Mains 2026 21st January Shift 2

Physics

The energy of an electron in an orbit of the Bohr's atom is

-0.04 \mathrm{E}_{0} \mathrm{eV}

where

\mathrm{E}_{0}

is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck's constant, then

\frac{2 \pi \mathrm{~L}}{\mathrm{~h}}

\_\_\_\_

Correct Answer:

Angular momentum

\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}\\

\mathrm{n}=\frac{2 \pi \mathrm{~L}}{\mathrm{~h}}\\

Energy

\mathrm{E}=-\frac{13.6}{\mathrm{n}^{2}} . \mathrm{z}^{2}

\\ \begin{aligned} & \mathrm{E} \Rightarrow-\frac{\mathrm{E}_{0}}{\mathrm{n}^{2}}=-0.04 \mathrm{E}_{0} \\ & \mathrm{n}^{2}=25, \mathrm{n}=5 \end{aligned}

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