Question

Calculate the longest and shortest wavelengths of the light emitted by electrons in the hydrogen atom that begin in the n=6 state and then fall to states with smaller values of N.

Answer 1

Answer: Longest wavelength = 7467.317806nm

Shortest wavelength = 93.8nm

Explanation:

For the Longest wavelength:

N=5

E= -R*(1/nf2 – 1/ni2)

R = 2.178*10^-18 J

Where E = (2.178*10^-18 J)*(1/52 – 1/62) = 2.662*10^-20J

Given that WL = hc/E

h = planck constant = 6.626*10^-34

c = speed of light = 3*10^8

WL = (6.626*10^-34)*(3*10^8) / (2.662*10^-20 ) = 7467.317806nm

For the Shortest wavelength:

N = 1

E= -R*(1/nf2 – 1/ni2)

R = 2.178*10^-18 J

Where E = (2.178*10^-18 J)*(1/12 – 1/62) = 2.1175*10^-18 J

Given that WL = hc/E

h = planck constant = 6.626*10^-34

c = speed of light = 3*10^8

WL = (6.626*10^-34)*(3*10^8) / (2.1175*10^-18 ) = 93.8nm

I hope this helps to explain the question.