The magnitude of total energy and angular mom | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) According to Bohr's model,the energy of an electron in the $n^{th}$ orbit is given by $E_{n} = -\frac{13.6}{n^{2}} 	ext{ eV}$.Thus,$E_{n} \propto

Vedclass · Accepted Answer

$E_{n} \propto \frac{1}{L_{n}^{2}}$

Vedclass · Answer

(C) According to Bohr's model,the energy of an electron in the $n^{th}$ orbit is given by $E_{n} = -\frac{13.6}{n^{2}} 	ext{ eV}$.Thus,$E_{n} \propto \frac{1}{n^{2}}$.The angular momentum of an electron in the $n^{th}$ orbit is given by $L_{n} = \frac{nh}{2\pi}$.Thus,$L_{n} \propto n$,which implies $n \propto L_{n}$.Substituting $n \propto L_{n}$ into the energy relation,we get $E_{n} \propto \frac{1}{L_{n}^{2}}$.Therefore,the correct relation is $E_{n} \propto \frac{1}{L_{n}^{2}}$.

The magnitude of total energy and angular momentum of an electron in the $n^{th}$ orbit of a Bohr atom is denoted by $E_{n}$ and $L_{n}$ respectively. Then

Similar Questions

The triply ionized beryllium $(Be^{3+})$ has the same electron orbital radius as that of the ground state of hydrogen. Hence, the energy state of triply ionized beryllium is (Given $Z = 4$ for beryllium)

When an electron in a Hydrogen atom revolves in a stationary orbit,it

In a hydrogen atom,if $V_n$ and $V_p$ are the orbital velocities in the $n^{\text{th}}$ and $p^{\text{th}}$ orbits respectively,then the ratio $V_p : V_n$ is:

The kinetic energy of the electron in an orbit of radius $r$ in a hydrogen atom is ($e =$ electronic charge).

The total energy of an electron in the $n^{th}$ orbit of a hydrogen atom is $E_n$. The total energy of an electron in the $n^{th}$ orbit of a $He^+$ ion is .........

Vedclass Test Series

Exam Paper Generator

Online Exam Module