The energy of an electron in the first Bohr orbit of the H-atom is -13.6 eV. The magnitude of energy value of an electron in the first excited state of Be3+ is eV (nearest integer value).
Answer
54
Solution
For a hydrogen-like species, En = -13.6 Z2n2 eV
For Be3+, Z = 4; first excited state n = 2
E2 = -13.6 4222 = -13.6 4 = -54.4 eV
Magnitude = 54.4 eV 54 eV
Answer: 54