Question

MODERN PHYSICS

A photon emitted from an excited hydrogen atom has an energy of 3.02 electron volts. Which electron energy-level transition would produce this photon?


A. n=1 to n=6

B. n=2 to n=6

C. n=6 to n=1

D. n=6 to n=2


I chose B but the correct answer is D can someone tell me why? And what's the difference?

Answer 1

D. n=6 to n=2

Step-by-step explanation:

Given;

energy of emitted photon, E = 3.02 electron volts

The energy levels of a Hydrogen atom is given as; E = -E₀ /n²

where;

E₀ is the energy level of an electron in ground state = -13.6 eV

n is the energy level

From the equation above make n, the subject of the formula;

n² = -E₀ / E

n² = 13.6 eV / 3.02 eV

n² = 4.5

n = √4.5

n = 2

When electron moves from higher energy level to a lower energy level it emits photons;

`E = E_0((1)/(n_1^2)-(1)/(n_2^2) )\\\\(1)/(n_1^2)-(1)/(n_2^2) = (E)/(E_o) \\\\(1)/(4) -(1)/(n_2^2) = (3.02)/(13.6) \\\\(1)/(4) -(1)/(n_2^2) =0.222\\\\(1)/(n_2^2) = 0.25 - 0.22\\\\(1)/(n_2^2) = 0.03\\\\n_2^2 = 33.33\\\\n_2 = √(33.33) = 6`

For a photon to be emitted, electron must move from higher energy level to a lower energy level. The higher energy level is 6 while the lower energy level is 2

Therefore, The electron energy-level transition is from n = 6 to n = 2