Question

A hydrogen atom with its electron in the n = 6 energy level emits a photon of IR light. Calculate (a) the change in energy of the atom and (b) the wavelength (in Å) of the photon.

Answer 1

Answer:

(a) The change in energy of the atom is --6.0528 X 10⁻²⁰ J

(b) The wavelength of the photon is 32841 Å

Explanation:

E = -kz²/n²

Where;

E is the change in the energy of the atom

k is a constant = 2.179 X 10¹⁸ J

z is the atomic number of hydrogen = 1

n = 6

E = -(1 X 2.179 X 10⁻¹⁸ )/6²

E = -6.0528 X 10⁻²⁰ J

Therefore,the change in energy of the atom is -6.0528 X 10⁻²⁰ J

(b) the wavelength (in Å) of the photon, can be calculated as follows;

E = hc/λ

λ = hc/E

where;

h is Planck's constant = 6.626 X 10⁻³⁴ js

c is the speed of light = 3 X 10⁸ m/s

λ  is the wavelength of the photon

λ  = (6.626 X 10⁻³⁴ X 3 X 10⁸ )/(6.0528 X 10⁻²⁰)

λ  = 3.2841 X 10⁻⁶ m

λ  = 32841 Å

Therefore, the wavelength of the photon is 32841 Å