Question

A ground state hydrogen atom absorbs a photon of light having a wavelength of 93.73 nm. It then gives off a photon having a wavelength of 2624 nm. What is the final state of the hydrogen atom?

Answer 1

1/λ = R[1/(n1^2) -1/(n2^2)]

Where λ = the wavelength in m of the light emitted (or absorbed) = 93.73×10^-9 m

R is the Rydberg constant; R = 1.09737×10^7 m-1

n1 and n2 are integers such that n1< n2; n1 = 1, n2 = to find First Part Absorption

1/93.73×10^-9 = 1.097×10^7 [1/1^1 -1/n2^2]

1 = 93.73×10^-9×1.09737×10^7[1-1/n2^2]

1 = 1.029 -1.029/n^2

1.029/n^2 = 0.029

n^2 = 1.029/0.029 = 35.5
n2 = 6

Emission n2 = 6 n1 = to find, λ = 2624×10^-9

1/2624×10^-9= 1.09737×10^7[1/n^2 -1/6^2]

1 = 2630×10^-9×1.09737×10^7[1/n^2-0.02777]

1 = 28.861/n^2 – 0.8017

28.861/n^2 = 1.8017

n^2 = 28.861/1.8017= 16.0

n1 = 4 The final state of the H atom

final state = 4