finding n=f for a ground state

Question

a ground state of hydrogen atom absorbs a photon of light having a wavelength of 92.27nm. it gives off a photon having a wavelength of 922.7nm. what is the final state of hydrogen atom? nf=

Answer 1

The change in energy for a hydrogen electron, ∆E, is given by the equation:

∆E = E_final - E_initial = R_H (1/n_f² - 1/n_i²)

E_final = final energy

E_initial = initial energy

R_H = -2.179x10^-18 J

nf = final energy level = ?

ni = initial energy level = 1 (ground state)

We can find the initial and final energies from the given wavelengths of the photons:

E = hc / λ where E = energy; h = Planks constant (6.626x10^-34 Js; c = speed of light (3x10⁸ m/s);

λ = wavelength in meters

For E_initial = (6.626x10^-34 Js)(3x10⁸ m/s) / 92.27x10^-9 m) = 2.15x10^-18 J

For E_final = (6.626x10^-34 Js)(3x10⁸ m/s) / 922.7x10^-9 m) = 2.15x10^-19 J

∆E = E_final - E_initial = -R_H (1/n_f² - 1/n_i²)

2.15x10^-19 J - 2.15x10^-18 J = 1.94x10^-18 J = - 2.179x10^-18 J (1/(nf)² - 1/1²)

1.94x10^-18 J = - 2.179x10^-18 J (1/(n_f)² - 1)

1.94x10^-18 J = -2.179x10^-18 / n_f² + 2.179x10^-18

-2.39x10^-19 = -2.179x10^-18 / n_f²

n_f = 3