Q113 / 150 · MHT CET 2026 (15 April Shift 1) · Physics
If 'E' and 'L' denote the magnitude of total energy and angular momentum of revolving electron in $n^{th}$ Bohr orbit then
- A$E \propto L$
- B$E \propto L^{-1}$
- Correct answerC$E \propto L^{-2}$
- D$E \propto L^{2}$
Solution
The magnitude of total energy of an electron in the $n^{th}$ Bohr orbit is given by $E \propto \dfrac{1}{n^2}$.
The angular momentum of the electron in the $n^{th}$ Bohr orbit is given by $L = \dfrac{nh}{2\pi}$.
This implies $L \propto n$.
Substituting $n \propto L$ in the expression for energy, we get $E \propto \dfrac{1}{L^2}$.
Thus, $E \propto L^{-2}$.
Answer: $E \propto L^{-2}$
The angular momentum of the electron in the $n^{th}$ Bohr orbit is given by $L = \dfrac{nh}{2\pi}$.
This implies $L \propto n$.
Substituting $n \propto L$ in the expression for energy, we get $E \propto \dfrac{1}{L^2}$.
Thus, $E \propto L^{-2}$.
Answer: $E \propto L^{-2}$
Hence, the correct answer is option: 'C'.