A student of physics says: “an electron falls from an energy level of -4.5 eV to -7.2 eV. It emits a photon with 2.7 eV of energy”. Give your comment on his claim.

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Energy of an electron in the n_th orbit is given by the energy expression

`E_n = - (2pi^2me^4Z^2)/(n^2h^2)`

This equation calculates the energy of an electron as a negative number. This is because a completely free electron (`n=oo` ) is assigned an energy of zero. When electrons become bound within atoms, they become stabilized through attraction of the positively charged nucleus and their energy falls. As it is falling from zero, it must be negative. This means that as an electron falls from higher orbit to lower ones, its energy has increasingly larger negative values relative to zero, the energy for the free electron. The lower orbits thus have more negative energies than the higher ones. The energies of orbits are related as

`E_n=E_0*n^2`

When the student says “an electron falls from an energy level of -4.5 eV to -7.2 eV”, several conclusions can be made out of it:

1. He must be referring to an atom other than hydrogen (because, E_0 for H-atom is -13.6 eV, and tha data does not fit into the formula `E_o/n^2` , with integer n).

2. The energy levels must be considerably higher numbered, otherwise it could not have assumed such a low negative value.

3. He is talking about transition from a higher energy level to a lower energy level.

4. Accordingly, there will be emission of photon whose energy, as per the law of conservation of energy, is equal to the difference in energy between these two energy levels.

Thus energy of the photon = `DeltaE= E_f-E_i` =7.2-4.5=2.7 eV.

5. A radiation of matching frequency `nu` such that `DeltaE=hnu` will be associated with this transition.

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