Correct Answer - Option 4 : The P.E. increases and K.E. decreases
Concept:
When an electron revolves around the nucleus, it carries Kinetic energy as well as Potential energy.
While Revolving around the nucleus, its kinetic energy is given by
\(K.E. = \dfrac{Z^2me^4}{8n^2h^2\epsilon_0}\)
and Potential Energy is given by
\(P.E. = -\dfrac{Z^2me^4}{4n^2h^2\epsilon_0}\)
Thus, P.E.= -2K.E.
Now, Total energy is the sum of total kinetic energy and total potential energy.
i.e, TE = PE + KE
Explanation:
We know that,
\(K.E. = \dfrac{Z^2me^4}{8n^2h^2\epsilon_0}\)
K.E. \(\propto \dfrac{1}{n^2}\)
So, as the hydrogen atom is raised from the ground state to an excited state, n increases.
Hence, Kinetic energy decreases.
Also, Potential energy is given by
\(P.E. = -\dfrac{Z^2me^4}{4n^2h^2\epsilon_0}\)
P.E.\(\propto -\dfrac{1}{n^2}\)
So, as the hydrogen atom is raised from the ground state to an excited state, n increases
This decreases the negative value of Potential energy, i.e PE becomes less negative.
Hence, Potential energy increases.
Option (4) is the correct answer.