An electron in the ground state goes to the first excited state in the hydrogen atom. Arrange the following energies in decreasing order

(A) Total energy in ground state
(B) Kinetic energy in ground state
(C) Potential energy in ground state
(D) Total energy in first excited state

Choose the correct answer from the options given below:

(B), (D), (A), (C)

The correct answer is Option (3) → (B), (D), (A), (C)

For hydrogen atom:

Total energy: $E_n = -\frac{13.6}{n^2} \, \text{eV}$

Kinetic energy: $K_n = -E_n = \frac{13.6}{n^2} \, \text{eV}$

Potential energy: $U_n = 2E_n = -\frac{27.2}{n^2} \, \text{eV}$

Ground state ($n=1$): $E_1 = -13.6 \, \text{eV}$, $K_1 = 13.6 \, \text{eV}$, $U_1 = -27.2 \, \text{eV}$

First excited state ($n=2$): $E_2 = -3.4 \, \text{eV}$

Decreasing order: $K_1 (13.6) > E_2 (-3.4) > E_1 (-13.6) > U_1 (-27.2)$

Decreasing order: B, D, A, C