Question

First excitation potential for the Hydrogen like (hypothetical) sample is $24$V. Choose the correct statement.
A. First excitation potential for the hydrogen like (hypothetical) sample is $24$V.
B. Ionisation energy of the sample is $32$eV.
C. Binding energy of the third excited state is $2$eV.
D. Second excitation potential of the sample is $\dfrac{32\times 8}{9}V$.

Answer 1

You should know that; the first excitation potential of a substance is the energy released by an electron when it drops from the first excited state to the ground state. The ionisation energy is the energy which is released when the electron comes to the ground state from any state. This problem can be solved by using the formula for the energy differences when the electron jumps between two excited states.

Complete step by step solution:
As explained in the hint, we will use the formula for the energy released by an electron in the hydrogen like sample when jumping between two different excited states to get first the first excitation potential and the ionisation energy. Then we can compare the correct statements.
The energy released (i.e. the excitation energy) in an atom when it jumps from ${{n}_{1}}$ excited state to ${{n}_{2}}$ ground state is given by:
$E\text{excitation energy}=-{{E}_{0}}\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]={{E}_{0}}\left[ \dfrac{1}{n_{2}^{2}}-\dfrac{1}{n_{1}^{2}} \right]$
Where, $-{{E}_{0}}$ is considered as the ionization energy.
So, here in the given question it is stated that first excitation potential for the Hydrogen like (hypothetical) sample is $24$V.
So, here ${{n}_{1}}$ will be $2$ and ${{n}_{2}}$ will be $1$ for the first excitation potential.
Therefore, the excitation potential will be $={{E}_{0}}\left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}} \right]=\dfrac{3}{4}{{E}_{0}}$
The excitation potential is already given as $24V$.
So, $24=\dfrac{3}{4}{{E}_{0}}$
Then, ${{E}_{0}}=\dfrac{24\times 4}{3}=32eV$
Similarly, for the second excitation potential (let it be E), ${{n}_{1}}$ will be $3$ and ${{n}_{2}}$ will be $1$.
So, by placing the value of the ionization energy in the formula mentioned above, we will get;
$E=32\left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{3}^{2}}} \right]=32\times \dfrac{8}{9}$
The binding energy (let it be B.E) can be calculated by dividing the ionisation potential to that of the square of the excited state and the formula is given as:
$B.E.=\dfrac{\text{Ionisation energy}}{{{n}^{2}}}$, where n is the excited state.
So, here it is given as the electron is present in the third excited state. So, n will be $4$.
Therefore, the binding energy for the third excited state will be
$B.E.=\dfrac{32}{{{4}^{2}}}=\dfrac{32}{16}=2eV$
By looking into each option, we can see that all the options are correct.

Hence, all the options are correct.

Note: It is important to be careful while writing the formula for the excitation state. The negative sign for the ionization energy holds a key role. It is due to this negative sign that the ultimate result drops out to be positive and signifies energy is released when an electron is jumped from the excited state to that of the ground state.