Question

In a transition to a state of excitation energy 10.19 eV, a hydrogen atom emits a 4890 Å photon. The binding energy of the initial state is

• A. 1.51 Ev
• B. 3.4eV
• C. 0.54 eV
• D. 0.87 eV

Answer 1

Answer: (D)

0.87 eV

Answer 2

Energy of emitted photon is

E = $\frac{hc}{\lambda}$= 2.54 eV

The excitation energy is the energy to excite the atom to a level above the ground state. Therefore the energy level is

E = – 13.6 + 10.19 = – 3.41 eV

Photon arises from transition between energy states such that

E_i – E_f = hn = 2.54 eV

E_i = 2.54 + E_f

E_i = 2.54 + (5) = 2.54 – 3.41 eV

E_i = – 0.87 Ev