The ratio of kinetic energy and the total energy of the electron in the nth quantum state of Bohr's atomic model of a hydrogen atom is

Question

The ratio of kinetic energy and the total energy of the electron in the nth quantum state of Bohr's atomic model of a hydrogen atom is (A) -2 (B) -1 (C) 2 (D) 1

Tardigrade Admin · Accepted Answer

The Kinetic energy of the electron in the nth state K=(m Z2 e4/8 ϵ 02 h2 n2) The total energy of the electron in the nth state T = -(m Z2 e4/8 ϵ 02 h2 n2) ∴ (K/T)= -1

Q. The ratio of kinetic energy and the total energy of the electron in the $n^{t h}$ quantum state of Bohr's atomic model of a hydrogen atom is

$- 2$

$- 1$

$2$

$1$

The Kinetic energy of the electron in the $n^{t h}$ state
$K = \frac{m Z ^{2} e ^{4}}{8 ϵ _{0}^{2} h ^{2} n ^{2}}$
The total energy of the electron in the $n^{t h}$ state
$T = - \frac{m Z ^{2} e ^{4}}{8 ϵ _{0}^{2} h ^{2} n ^{2}}$
$∴ \frac{K}{T} = - 1$

Q. The ratio of kinetic energy and the total energy of the electron in the nth quantum state of Bohr's atomic model of a hydrogen atom is

−2

−1

2

1

The Kinetic energy of the electron in the nth state K=8ϵ02​h2n2mZ2e4​ The total energy of the electron in the nth state T=−8ϵ02​h2n2mZ2e4​ ∴TK​=−1

Q. The ratio of kinetic energy and the total energy of the electron in the $n^{t h}$ quantum state of Bohr's atomic model of a hydrogen atom is

$- 2$

$- 1$

$2$

$1$

The Kinetic energy of the electron in the $n^{t h}$ state
$K = \frac{m Z ^{2} e ^{4}}{8 ϵ _{0}^{2} h ^{2} n ^{2}}$
The total energy of the electron in the $n^{t h}$ state
$T = - \frac{m Z ^{2} e ^{4}}{8 ϵ _{0}^{2} h ^{2} n ^{2}}$
$∴ \frac{K}{T} = - 1$