An electron transition takes place from excited state to ground state in hydrogen atom, then

The total energy of an electron in the \(n\)-th orbit of a hydrogen atom is given by \(E = -\dfrac{13.6}{n^2}\) eV.

The kinetic energy is \(K = -E = \dfrac{13.6}{n^2}\) eV.

The potential energy is \(U = 2E = -\dfrac{27.2}{n^2}\) eV.

When the electron transitions from an excited state to the ground state, the principal quantum number \(n\) decreases.

As \(n\) decreases, the value of \(\dfrac{1}{n^2}\) increases. Therefore, the kinetic energy \(K\) increases.

Since the total energy \(E\) and potential energy \(U\) are negative and their magnitudes increase, their actual values become more negative. Thus, both potential energy and total energy decrease.

Answer: Its kinetic energy increases but potential energy and total energy decrease