When a hydrogen atom is raised from ground energy level to excited energy level, then:
(A) Potential energy increased and kinetic energy decreases
(B) Kinetic energy increases and Potential energy decreased
(c ) Both KE and PE increases
(D) Both KE and PE decreases

Hint:
When a hydrogen atom is raised from ground energy to excited energy heat must be applied.
Now, as it moves from lower to higher energy, the distance between the nucleus and electron increases. Potential energy is dependent on this distance.
Moreover, total energy is always constant.

Formula used:
1. \[PE = \dfrac{{ - KZ{e^2}}}{r}\]

Where,
\[PE = \] Potential Energy
\[K = \]Boltzmann constant
\[Z = \]Atomic shell
\[r = \] Radius of the orbit
\[e = \] Charge of electron

2. \[KE = \dfrac{{KZ{e^2}}}{{{r^2}}}\]

Where,
\[KE = \]Kinetic Energy
\[K = \]Boltzmann constant
\[Z = \]Atomic shell
\[r = \] Radius of the orbit
\[e = \] Charge of electron

Complete Step By Step Answer:
When an electron is excited to jump from ground level to higher energy level, energy is supplied.
Now, as it goes to a higher energy level, the radius of the orbit increases.
Now, let us consider the formula for Kinetic Energy:

 \[KE = \dfrac{{KZ{e^2}}}{{{r^2}}}\]

Where,
\[KE = \]Kinetic Energy
\[K = \]Boltzmann constant
\[Z = \]Atomic shell
\[r = \] Radius of the orbit
\[e = \] Charge of electron
When the radius \[r\]increases as the electron jumps from lower to higher energy level, kinetic Energy which is inversely proportional to square of the radius, Kinetic energy decreases.
Therefore, the more increase in \[r\] the lesser the kinetic energy becomes.
Now, in case of potential energy, we know:

\[PE = \dfrac{{ - KZ{e^2}}}{r}\]

Where,
\[PE = \] Potential Energy
\[K = \]Boltzmann constant
\[Z = \]Atomic shell
\[r = \] Radius of the orbit
\[e = \] Charge of electron

So, Option (A) Potential energy increased and kinetic energy decreases is correct.

Note:Here, it is seen that potential energy also varies inversely with\[r\], but due to that negative sign, the net potential energy increases.

Moreover, in case of kinetic energy, KE is inversely proportional to square off\[r\], whereas, PE is inversely proportional to \[r\] linearly. Thus KE is decreased more, whereas there is an increase in PE.