Question

A hydrogen atom in a state having a binding energy (the energy required to remove an electron) of -1.51 eV makes a transition to a state with an excitation energy (the difference between the energy of the state and that of the ground state) of 0 eV.(a) What is the energy of the photon emitted as a result of the transition?

(b) What is the higher quantum number of the transition producing this emission?
(c) What is the lower quantum number of the transition producing this emission?

Answer 1

a) ΔE = 1.5 eV , b) n=∞, c) n= 3

Step-by-step explanation:

a) the energy of the transition is the energy difference between the two states

ΔE = E₂ –E₁

ΔE = 0 - (-1.5)

ΔE = 1.5 eV

b) The energy of each state is given by the equation

E = -13.606 / n²

The highest quantum number is

n = √ (13,606 / E)

n = √ (13.606 /1.5)

n = 3

the transition is n = ∞ that has energy 0

c) the lowest quantum number is 3