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Question: If the first excitation potential of a hydrogen atom is \('V'\) electron volt, then the ionization e...

If the first excitation potential of a hydrogen atom is ′V′'V' electron volt, then the ionization energy of this atom will be
(a). VV electron volt
(b). 3V4\dfrac{3V}{4} electron volt
(c). 4V3\dfrac{4V}{3} electron volt
(d). Cannot be calculated by given information

Explanation

Solution

Hint: The first excitation potential of a hydrogen is the energy released by an electron when it drops from the first excited state to the ground state in the hydrogen atom. The ionization energy is the energy released when the electron comes to the ground state in the hydrogen atom from infinity. This problem can be solved by using the formula for the energy difference when the electron jumps between two different excited states.

Formula used:
The energy released EE by an electron in a hydrogen atom when it jumps from n2{{n}_{2}} excited state to n1{{n}_{1}} excited state is given by
E=−E0(1n12−1n22)E=-{{E}_{0}}\left( \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}} \right)
where E0=13.6eV{{E}_{0}}=13.6eV.

Complete step by step answer:
As explained in the hint, we will use the formula for the energy released by an electron in a hydrogen atom when jumping between different excited states to get the first excitation potential and the ionization energy. Then we can compare both these values to get the required answer.
The energy released EE by an electron in a hydrogen atom when it jumps from n2{{n}_{2}} excited state to n1{{n}_{1}} excited state is given by
E=−E0(1n12−1n22)E=-{{E}_{0}}\left( \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}} \right) --(1)
where E0=13.6eV{{E}_{0}}=13.6eV.
The first excitation potential VV is the energy released when the electron drops from the first excited state (n1=2)\left( {{n}_{1}}=2 \right) to the ground state (n2=1)\left( {{n}_{2}}=1 \right). Hence, using (1), we get,
V=−E0(122−112)=−E0(14−11)=−E0(1−44)=−E0(−34)=34E0V=-{{E}_{0}}\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{1}^{2}}} \right)=-{{E}_{0}}\left( \dfrac{1}{4}-\dfrac{1}{1} \right)=-{{E}_{0}}\left( \dfrac{1-4}{4} \right)=-{{E}_{0}}\left( \dfrac{-3}{4} \right)=\dfrac{3}{4}{{E}_{0}}
∴E0=43V\therefore {{E}_{0}}=\dfrac{4}{3}V --(2)
The ionization energy (I.E.)\left( I.E. \right) of the electron in a hydrogen atom is the energy released by the electron when the electron drops from infinity (n1=∞)\left( {{n}_{1}}=\infty \right) to the ground state (n2=1)\left( {{n}_{2}}=1 \right) . Hence, using (1), we get,
I.E.=−E0(1∞2−112)=−E0(0−1)=−E0(−1)=E0I.E.=-{{E}_{0}}\left( \dfrac{1}{{{\infty }^{2}}}-\dfrac{1}{{{1}^{2}}} \right)=-{{E}_{0}}\left( 0-1 \right)=-{{E}_{0}}\left( -1 \right)={{E}_{0}} (∵1∞=0)\left( \because \dfrac{1}{\infty }=0 \right)
∴IE=43V\therefore IE=\dfrac{4}{3}V [Using(2)]
Hence, we get the required ionization energy value as 4V3\dfrac{4V}{3}.
Therefore, the correct answer is C) 4V3\dfrac{4V}{3} electron volt.

Note: Students must be careful while writing the formula for equation (1), since the negative sign holds importance. It is due to the negative sign that the ultimate result comes out to be positive and this signifies that energy is released when an electron drops from a higher excited state to a lower excited state. Often students make this mistake and are not consistent with the formula which leads to errors and wrong interpretations of the energy change in these types of questions.