what is the energy of the photon emitted or absorbed?

Question

If the electron in singly ionized helium transitions from n = 1 to n = 3, what is the energy of the photon emitted or absorbed?

a) 54.4 eV

b) 12.1 eV

c) 9.1 eV

d) 48.4 eV

e) 6.04 eV

Correct answer is D

Answer 1

At n = 1 (ground state), the energy level for singly ionized He is

E₁ = Z²(-13.6 eV) / n²

Z = 2

n = 1

E₁ = 2²(-13.6 eV) / 1² = -54.4 eV

Similarly, at n = 3, the energy level for singly ionized He is

E₃ = 2²(-13.6 eV) / 3² = -6.04 eV

E₃ - E₁ = [-6.04 - (-54.4)] eV = 48.4 eV