Question

A hydrogen electron in the ground state absorbs a 92.3157 nm photon to reach a higher excited state (a), and then emits two photons, one with 1.75777 x 10^-20 J of energy to reach an intermediate state (b), and one with wavelength 1005.22nm as it falls back to a lower excited state (c). Determine the three energy levels (a,b,c) of this particular electron.

Answer 1

We know that

Ground state energy of hydrogen atoms = -13.6eV

And we also know that 1eV = 1.602×10^-19J

So energy of ground state = -13.6×1.602×10^-19J = - 21.7872×10^-19 J

Energy of ground state = -2.17872×10^-18J

Step(1)

Energy of state (a) = energy of ground state + energy of absorbed photon

Energy of photos = hC/

= (6.626×10^-34.J-s)(3×10⁸m/s)/(92.3157×10^-9m)

Energy of photos = 2.15326×10^-18J

So

Energy of state (a) = -2.17872×10^-18 J + 2.25326×10^-18 J

=-0.02546×10^-18J = -2.546×10^-20J

^step(2)

Energy of state (b) = energy of state (a) - energy of emitted photon

Energy of state (b) = -2.546×10^-20J - 1.75777×10^-20J

Energy of state (b) = - 4.30377×10^-20J

Step(3)

Energy of state (c) = energy of state (b) - energy of emitted photon

Energy of emitted photon = hC/

= (6.626×10^-34J-s)(3×10⁸m/s)/(1005.22×10^-9m)

Energy of emitted photon = 0.019774×10^-17J = 19.774×10^-20 J

Energy of state (c) = -4.30377×10^-20 J - 19.774×10^-20 J

= 24.07854×10^-20J

Energy of state (c) = 2.407854×10^-19J