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(C) The energy of the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -\frac{13.6 Z^2}{n^2} \, 	ext{eV}$.The first excitation potential cor

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$\frac{4V}{3} \, 	ext{eV}$

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(C) The energy of the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -\frac{13.6 Z^2}{n^2} \, 	ext{eV}$.The first excitation potential corresponds to the energy required to excite an electron from the ground state $(n=1)$ to the first excited state $(n=2)$.The energy difference is $\Delta E = E_2 - E_1 = 13.6 Z^2 \left( 1 - \frac{1}{4} ight) = 13.6 Z^2 \left( \frac{3}{4} ight)$.Given that the first excitation potential is $V$, we have $V = \frac{3}{4} E_i$, where $E_i = 13.6 Z^2$ is the ionization energy.Therefore, the ionization energy $E_i = \frac{4}{3} V \, 	ext{eV}$.

If the first excitation potential of a hydrogen-like atom is $V \, \text{volt}$, then the ionization energy of this atom will be

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