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Question: Out of the following which one is not a possible energy for a photon to be emitted by a hydrogen ato...

Out of the following which one is not a possible energy for a photon to be emitted by a hydrogen atom according to Bohr’s atomic model?
(a) 1⋅9eV1\cdot 9eV
(b) 11⋅1eV11\cdot 1eV
(c) 13⋅6eV13\cdot 6eV
(d) 0⋅65eV0\cdot 65eV

Explanation

Solution

The difference in Bohr orbital energies for H-atoms gives the energy emitted by the photons and hence we can choose which one is the correct option or not.
The energy is expressed as a negative number because it takes that much energy to unbind the electron from the nucleus. If a photon has more energy than the binding energy of the electron then the photon will free the electron from the atom-ionizing it. The ground state is the most bound state and therefore takes the most energy to ionize.

Complete step by step answer:
The energy of the nth{{n}^{th}} orbital of H-atom is given by:
En=−13⋅6n2eV{{E}_{n}}=\dfrac{-13\cdot 6}{{{n}^{2}}}eV
So,
E1=−13⋅6(1)2=−13⋅6eV{{E}_{1}}=\dfrac{-13\cdot 6}{{{(1)}^{2}}}=-13\cdot 6eV
E2=−13⋅6(2)2=−3⋅4eV{{E}_{2}}=\dfrac{-13\cdot 6}{{{(2)}^{2}}}=-3\cdot 4eV
E3=−13⋅6(3)2=−1⋅5eV{{E}_{3}}=\dfrac{-13\cdot 6}{{{(3)}^{2}}}=-1\cdot 5eV
E4=−13⋅6(4)2=−0⋅85eV{{E}_{4}}=\dfrac{-13\cdot 6}{{{(4)}^{2}}}=-0\cdot 85eV
Energy of photons emitted is given by the energy difference between the two energy levels between which the photon emission has taken place. According to Bohr’s atomic model when photon is emitted from 3rd orbit to the 2nd orbit, energy of photon is given by:
E3−E2=[−1⋅5−(−3⋅4)]eV{{E}_{3}}-{{E}_{2}}=\left[ -1\cdot 5-\left( -3\cdot 4 \right) \right]eV
=[−1⋅5+3⋅4]eV=\left[ -1\cdot 5+3\cdot 4 \right]eV
=1⋅9eV=1\cdot 9eV
Similarly, energy of photon emitted from 4th orbit to 3rd orbit is given by:
E4−E3=[−0⋅85−(−1⋅5)]eV{{E}_{4}}-{{E}_{3}}=\left[ -0\cdot 85-\left( -1\cdot 5 \right) \right]eV
=[−0⋅85+1⋅5]eV=\left[ -0\cdot 85+1\cdot 5 \right]eV
=0⋅65eV=0\cdot 65eV
Photon emitted from nth level with energy = 0 eV to the 1st orbit has energy given by:
En−E1=[0−(−13⋅6)]eV =13⋅6eV \begin{aligned} & {{E}_{n}}-{{E}_{1}}=\left[ 0-\left( -13\cdot 6 \right) \right]eV \\\ & =13\cdot 6eV \\\ \end{aligned}
Therefore, 11⋅111\cdot 1 eV is the energy which is not possible to be the energy of the emitted photon.

Note:
While calculating this should be kept in mind that the energy of the nth^{\text{th}} orbital electron is not the same as the energy of the photon emitted by the orbital. From the above formula we can find out the energy of an atom and orbital both. Bohr's model is simpler and relatively easy to understand. It is useful because it helps us to understand what's observed in nature.Bohr's theory does not work for helium. It could not explain properties of atoms having more than one electron. Helium has two electrons that is why it does not apply on helium or any atoms which have more than one electron.