Atomic Energy Levels. An atom in the second excited state (n = 3) of H is...

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.)

Expert Solution

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

genius_generous answered 2 years ago

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