Question

When an electron in a hydrogen atom is raised from the ground state to an excited state, then,

• A. its P.E. increases and K.E. decreases.
• B. its P.E. decreases and K.E. increases.
• C. Both K.E and P.E increase.
• D. Both K.E and P.E decrease.

Answer 1

The correct option is A its P.E. increases and K.E. decreases.

For an electron in a hydrogen atom,

$K.E.=\frac{m{e}^4}{8{ϵ}_0^2{h}^2{n}^2}$

$P.E.=\frac{-m{e}^4}{4{ϵ}_0^2{h}^2{n}^2}$

As the electron goes in excited state, $n$ increases so that, K.E. decreases and P.E. increases.

Hence, $\left(A\right)$ is the correct answer.

$$\begin{array}{c}\text{Why this Question ?}\\ \text{The question will have a clear understanding on Bohr's third postulate which is based on energy in different orbits.}\end{array}$$