JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Atomic Structure

The energy of an electron in first Bohr orbit of H-atom is -13.6 eV. The magnitude of energy value of electron in the first excited state of

Be^{3 +}

is _____ eV. (nearest integer value)

Answer: 54

Solution

For a hydrogen-like species the orbital energy is

E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}} eV

. For

B e^{3 +}

the nuclear charge is

Z = 4

, and the first excited state is

n = 2

, so

E_{2} = - 13.6 \times \frac{4 ^{2}}{2 ^{2}} = - 13.6 \times 4 = - 54.4 eV

. The magnitude of the energy of the electron in this state is

54 eV

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Integer answer