Question

After a certain H atom emits a 3740 nm photon, the energy of its electron is -8.716 x 10-20 J. What was the electron's energy before photon emission, and what level did it occupy?

Answer 1

The energy of the photon emitted during transition

ΔE = h ( c / λ)

where ΔE represents the change in energy of the atom, which equals the energy of the emitted photon, h is Planck's costant which is 6.626x10-34 J·s , c is the speed of light in a vacuum is 3x10^8 m/s and λ is the wavelength in meters.

λ = 3740 nm
1nm = 10^-9 m
λ = 3740 nm x (10^-9 m / 1 nm) = 3.74x10^-6 m

ΔE = h ( c / λ)

ΔE = (6.626x10-34 J·s)(3x10^8 m/s / 3.74^-6 m)

ΔE = 5.313 x 10^-20 J
(Since energy is absorbed during transition, the sign of ΔE is positive)

ΔE = energy of final state - energy of initial state

For n = 5
E5 = - 2.18x10^-18 J (1^2 / 5^2) = - 8.7x10^-20 J= hence,final state is n=5
now ,we need to find initial state

ΔE = E5 - En

En = ΔE - E5

En = 5.313 x 10^-20 J - (- 8.716 x 10-20 J) = 14.029x10^-20 J

En = - 2.18x10^-18 J (1^2 / n^2) = -14.029x10^-20 J

(1^2 / n^2) = -14.029x10^-20 / - 2.18x10^-18 J = 0.0645

n^2 = 1 / 0.0645 = 15.504
n = 3.93 ≈ 4

electron energy before transmission=-14.029x10^-20 J

level it occupy=4