Question

The ratio of kinetic energy and potential energy of an electron in a Bohr-orbit of a hydrogen like series is:
A) $\dfrac{1}{2}$.
B) $-\dfrac{1}{2}$.
C) 1
D) -1

Answer 1

The total energy is the sum of kinetic and potential energy. By using the virial theorem the relation between total energy and potential energy is calculated. Similarly, the relation between total energy and kinetic energy is calculated. By substituting the value of total energy from Bohr Theory and then by comparing the kinetic energy and potential energy the ratio can be calculated.

Complete answer:
The relation between kinetic energy and potential energy is given by virial theorem.
According to the virial theorem, for the hydrogen atom the relation between kinetic and potential energy is as follows:
${\text{2T}} = \, - {\text{V}}$
Where,
T is kinetic energy.
V is the potential energy.
The total energy of the hydrogen atom is given as follows:
${\text{E}}\,{\text{ = }}\,{\text{T + }}\,{\text{V}}$….$(1)$
Where,
E is the total energy.
Rearrange the virial theorem for kinetics energy as follow:
${\text{T}} = \, - \dfrac{{\text{V}}}{2}$……(2)
Substitute the value of kinetics energy from equation (2) in energy of hydrogen atom in equation (1).
${\text{E}}\,{\text{ = }}\, - \dfrac{{\text{V}}}{2}{\text{ + }}\,{\text{V}}$
${\text{E}}\,{\text{ = }}\,\dfrac{{\text{V}}}{2}$
${\text{V = }}\,{\text{2E}}\,$…..${\text{(3)}}$
Rearrange the virial theorem for potential energy as follow:
${\text{V = }} - 2{\text{T}}$……(4)
Substitute the value of potential energy from equation (4) in energy of hydrogen atom in equation (1).
${\text{E}}\,{\text{ = }}\,{\text{T + }}\,\left( { - 2{\text{T}}} \right)$
${\text{E}}\,{\text{ = }}\, - {\text{T}}$…..${\text{(5)}}$
According to Bohr Theory, the energy of the hydrogen atom is,
${\text{E = }}\, - 13.6\,\dfrac{{{{\text{Z}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}}$ ….. (6)
Where,
Z is the atomic number.
n is the principal quantum number.
Put the value of energy from equation (6) into equation ${\text{(3)}}$.
${\text{V = }}\,{\text{2}}\left( { - 13.6\,\dfrac{{{{\text{Z}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}}} \right)\,$ ……${\text{(7)}}$
Put the value of energy from equation (6) into equation (5).
${\text{T}}\,{\text{ = }}\, - \left( { - 13.6\,\dfrac{{{{\text{Z}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}}} \right)\,$
${\text{T}}\,{\text{ = }}\,13.6\,\dfrac{{{{\text{Z}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}}$…..(8)
Take the ratio of reaction (8) and ${\text{(7)}}$.
$\dfrac{{{\text{T}}\,}}{{\text{V}}}{\text{ = }}\dfrac{{\,13.6\,\dfrac{{{{\text{Z}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}}}}{{\,{\text{2}}\left( { - 13.6\,\dfrac{{{{\text{Z}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}}} \right)\,}}$
$\dfrac{{{\text{T}}\,}}{{\text{V}}}{\text{ = }} - \dfrac{1}{2}$
So, the ratio of kinetic energy and potential energy of an electron in a Bohr-orbit of a hydrogen-like series is $- \dfrac{1}{2}$.

Therefore, option (B) $- \dfrac{1}{2}$ is correct.

Note: Total energy of the hydrogen atom is the sum of the kinetic and potential energy. According to the virial theorem, the kinetic energy is half of the negative of potential energy. The potential energy is the double of the negative of kinetic energy. The total energy is equal to the negative of the kinetic energy. The total energy is equal to half of the potential energy.