Question

Atomic Energy Levels. An atom in the second excited state (n = 3) of H is just barely ionized when a photon strikes the atom—what is the wavelength of this photon? (Assume that all of the photon’s energy went into the atom.)

Answer 1

The striking photon causes the electron in the third excited state to escape from the atom or raise to the n = infinity, as the atom get ionised.

Now, energy of the electron in n= 3 state is

E3 = -(13.6/3^2) eV = -1.51 eV

= -(1.51*1.6*10^-19) J = -2.416*10^-19 J

Energy of the electron in n = infinity, E' = 0

So, energy imparted by the photon in raising the electron from n = 3 to n = infinity is

E' - E3 = [0 - (-2.416*10^-19)] J = 2.416*10^-19 J

But, Energy = hf = hc/ , where h = 6.6*10^-34 Js

So, wavelength, = hc/Energy

Hence, = (6.6*10^-34*3*10^8/2.416*10^-19) m

= 8.20*10^-5 m