The magnitude of total energy and angular momentum of an electron in the $\mathrm{n}^{\mathrm{th}}$ orbit of…

The magnitude of total energy and angular momentum of an electron in the $\mathrm{n}^{\mathrm{th}}$ orbit of a Bohr atom is denoted by $E_{n}$ and $L_{n}$ respectively. Then
  1. $\mathrm{E}_{\mathrm{n}} \propto \mathrm{L}_{\mathrm{n}}$
  2. $\mathrm{E}_{\mathrm{n}} \propto \mathrm{L}_{\mathrm{n}}^{3}$
  3. $\mathrm{E}_{\mathrm{n}} \propto \frac{1}{\mathrm{~L}_{\mathrm{n}}^{2}}$
  4. $\mathrm{E}_{\mathrm{n}} \propto \frac{1}{\mathrm{~L}_{\mathrm{n}}}$

Solution

Given that, The total energy of the electron in the nth Bohr orbit is $E$. The angular momentum of an electron in the $n$th Bohr orbit is $L_n$ According to the Bohr theory, When the electron is in the nth Bohr orbit, then the total kinetic energy is inversely proportional to the square of the distance between them. That is given by, $E \propto \frac{1}{n^2}$......(1) And then, the linear momentum of the electron is inversely proportional to the distance between them. And the angular momentum of the electron is directly proportional to the distance between them. Here, the angular momentum is required, so the angular momentum of the electron is directly proportional to the distance between them is given by, $L_n \propto n$.....(2) By substituting the equation (2) in the equation (1), then the equation (1) is written as, $E \propto \frac{1}{L_n^2}$ Thus, the above equation shows the relation between the energy of the electron and the angular momentum of the electron in the nth Bohr orbit. :

Asked in: MHT CET 2020 (15 Oct Shift 2)

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