Problem 5: A photon is emitted from a hydrogen atom which undergoes a transition from that n = 3 state to the n = 2 state. Calculate (a) the energy (b) the wavelength, and (c) frequency of the emitted photon.
Solution
An electron in the higher energy level n = 3 has more energy than when it is in the lower energy level n =2. Therefore, when it transits from n = 3 to n = 2, the excess energy is emitted in the form of a photon (EM radiation). We have to calculate, the energy, wavelength and frequency of that emitted photon.
(a) We know that the energy of an electron in the energy level n is given by,
So we calculate the energy of electron when it is in n = 3 and then n = 2. For n = 3,
In the energy state n = 2, its energy is
Now the difference of the energies is the energy of the photon emitted. Therefore,
(b) We know that E = hf = hc/λ ⇒ λ = hc/E. Now h = 6.63 × 10-34 J s, c = 3 × 108 ms-1. Hence put the values; remembering the energy to be in Joules and 1 eV = 1.6 × 10-19 J.
(c) To find frequency, we use the formula, f = c/λ ; remembering the wavelength to be in meters. Put value of λ from the above equation
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