Question

A ground state hydrogen atom absorbs a photon of light having a wavelength of 92.27 nm. It then gives off a photon having a wavelength of 383.4 nm. What is the final state of the hydrogen atom? Values for physical constants can be found here. Number

Answer 1

step 1: calculate the final state after excitation

wavelength = 92.27 nm

wavelength = 9.227*10^-8 m

Here photon will be captured and it will excite the atom

1/lambda = -R* (1/nf^2 - 1/ni^2)

R is Rydberg constant. R = 1.097*10^7

1/lambda = -R* (1/nf^2 - 1/ni^2)

1/9.227*10^-8 = - 1.097*10^7*(1/nf^2 - 1/1^2)

(1/nf^2 - 1/1^2) = -0.9879

1/nf^2 = 1.205*10^-2

nf^2 = 83

nf = 9

step 2: Now calculate the final state after emission

wavelength = 383.4 nm

wavelength = 3.834*10^-7 m

Here photon will be captured and it will excite the atom

1/lambda = R* (1/nf^2 - 1/ni^2)

R is Rydberg constant. R = 1.097*10^7

1/lambda = R* (1/nf^2 - 1/ni^2)

1/3.834*10^-7 = 1.097*10^7*(1/nf^2 - 1/9^2)

(1/nf^2 - 1/9^2) = 0.2378

1/nf^2 = 0.2501

nf^2 = 4

nf = 2

Answer: 2