Question

An electron (in a hydrogen atom) in the n=5 state drops to the n=2 state by undergoing two successive downward jumps. What are all possible combinations of the resulting photon wavelengths?

Answer 1

all possibilities are

n = 5 ---> n = 4
n = 5 ---> n = 3
n = 5 --> n = 2
n = 4 ---> n = 3
n = 4 ---> n = 2
n = 3 ---> n = 2

1) n1 = 5 , n2 = 4

Energy of emited photon, E = 13.6eV*(1/n2^2 - 1/n1^2)

= 13.6*1.6*10^-19*(1/4^2 - 1/5^2)

= 4.896*10^-20 J

now use, E = h*c/lamda

==> lamda = h*c/E

= 6.626*10^-34*3*10^8/(4.896*10^-20)

= 4.06*10^-6 m

2) n1 = 5 , n2 = 3

Energy of emited photon, E = 13.6eV*(1/n2^2 - 1/n1^2)

= 13.6*1.6*10^-19*(1/3^2 - 1/5^2)

= 1.547*10^-19 J

now use, E = h*c/lamda

==> lamda = h*c/E

= 6.626*10^-34*3*10^8/(1.547*10^-19)

= 1.28*10^-6 m

3) n1 = 5 , n2 = 2

Energy of emited photon, E = 13.6eV*(1/n2^2 - 1/n1^2)

= 13.6*1.6*10^-19*(1/2^2 - 1/5^2)

= 4.57*10^-19 J

now use, E = h*c/lamda

==> lamda = h*c/E

= 6.626*10^-34*3*10^8/(4.57*10^-19)

= 4.35*10^-7 m 435 nm

4) n1 = 4 , n2 = 3

Energy of emited photon, E = 13.6eV*(1/n2^2 - 1/n1^2)

= 13.6*1.6*10^-19*(1/3^2 - 1/4^2)

= 1.06*10^-19 J

now use, E = h*c/lamda

==> lamda = h*c/E

= 6.626*10^-34*3*10^8/(1.06*10^-19)

= 1.88*10^-6 m

5) n1 = 4 , n2 = 2

Energy of emited photon, E = 13.6eV*(1/n2^2 - 1/n1^2)

= 13.6*1.6*10^-19*(1/2^2 - 1/4^2)

= 4.08*10^-19 J

now use, E = h*c/lamda

==> lamda = h*c/E

= 6.626*10^-34*3*10^8/(4.08*10^-19)

= 4.87*10^-7 m

6) n1 = 3 , n2 = 2

Energy of emited photon, E = 13.6eV*(1/n2^2 - 1/n1^2)

= 13.6*1.6*10^-19*(1/2^2 - 1/3^2)

= 3.02*10^-19 J

now use, E = h*c/lamda

==> lamda = h*c/E

= 6.626*10^-34*3*10^8/(3.02*10^-19)

= 6.58*10^-7 m