Question

Potential energy of electron present in \[H{e^ + }\] is:
(A) $\dfrac{{{e^2}}}{{2\pi {\varepsilon _o}r}}$
(B) $\dfrac{{3{e^2}}}{{4\pi {\varepsilon _o}r}}$
(C) $\dfrac{{ - 2{e^2}}}{{4\pi {\varepsilon _o}r}}$
(D) $\dfrac{{ - {e^2}}}{{4\pi {\varepsilon _o}r}}$

Answer 1

Hint:In order to solve these types of question we need to stuck in the concept which was known by us about the potential energy and use it as the question is telling to use not much is needed the need is only the potential energy formula for the orbital case which is $\dfrac{{ - Z{e^2}}}{{4\pi {\varepsilon _o}r}}$ .

Complete answer:For solving this we first need to know what is potential energy so we will start with potential energy:
It is defined as mechanical energy, stored energy, or energy caused by its position. The energy that a ball has when perched at the top of a steep hill while it is about to roll down is an example of the potential energy.So by the analysis of the gravitational potential energy we already calculated a general method for finding the potential energy by the electron which is
$PE = \dfrac{{ - KZee}}{r}$
On further solving
$PE = \dfrac{{ - KZ{e^2}}}{r}$
And finally after finding appropriate proportionality constant we get:
$PE = \dfrac{{ - Z{e^2}}}{{4\pi {\varepsilon _o}r}}$ ………………….(1)
Where,Z is atomic number of that element,
e is mass of electron,
r is radius of the electron,
As it is a helium atom so we know that its atomic number is 2.
Putting the value of Z in equation in equation (1) we get,
$PE = \dfrac{{ - 2{e^2}}}{{4\pi {\varepsilon _o}r}}$

So the correct option will be (C).

Note:Potential energy can also be defined as the negative of the work done by the internal forces.Derivation of gravitational potential energy;
$W = \int\limits_\infty ^r {\dfrac{{GMm}}{{{x^2}}}dx} $
On integrating we get
$W = - \left[ {\dfrac{{GMm}}{x}} \right]_\infty ^r$
Putting the upper and lower limits we get
$W = - \left[ {\dfrac{{GMm}}{r}} \right] - \left[ {\dfrac{{ - GMm}}{\infty }} \right]$
Finally we get
$W = \dfrac{{ - GMm}}{r}$