Problem 9: An electron drops from the second energy level to the first energy level within an excited hydrogen atom. (a) Determine the energy of the photon emitted. (b) calculate the frequency of the photon emitted (c) calculate the wavelength of the photon emitted.

Solution

Energy possessed by electron in energy level 2 is higher than its energy level 1. When it falls from second energy level to first, the excess of energy is emitted as photon of energy equal to the difference of energies in the two states. So we have to find the energies of the electron in the second and first levels and then calculate their difference. This will be the photon energy. In part (b) and (c) we will calculate the frequency and wavelength of this photon, respectively.

(a) We know that energy of an electron in the nth energy level is

Where E_{0} = 13.6 eV

Therefore, energy of the electron in n = 1,

The negative sign indicates the electron is bound to the atom.

Energy of electron in n = 2,

The difference in energies,

ΔE =-3.4 – (-13.6) = -3.4 + 13.6 = 10.2 eV

This gives the energy of the photon.

(b) Energy and frequency are related as, E = hf ⇒ f = E/h. where h = 6.63 × 10^{-34} J s. however, we first convert the unit of energy to joules. 1 e V = 1.6 × 10^{-19} J ⇒ 10.2 eV = 1.6 × 10^{-19} × 10.2 J. Now put these value in the equation,

This is the frequency of the emitted photon.

(c) To find wavelength, we use the relation, λ = c/f. Here c is the velocity of light = 3 × 10^{8} m/s. Put these values in the equation

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