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(D) Let the ground state energy of the hydrogen-like atom be $-E$.For the first excitation,the electron transitions from the ground state $(n=1)$ to t

Vedclass · Accepted Answer

$\frac{75}{4} \ V$

Vedclass · Answer

(D) Let the ground state energy of the hydrogen-like atom be $-E$.For the first excitation,the electron transitions from the ground state $(n=1)$ to the first excited state $(n=2)$.The energy required is $E_{2} - E_{1} = 15 \ eV$.$\left(\frac{-E}{2^2}ight) - \left(\frac{-E}{1^2}ight) = 15 \ eV$.$\frac{-E}{4} + E = 15 \ eV$.$\frac{3E}{4} = 15 \ eV \implies E = 20 \ eV$.For the third excitation,the electron transitions from the ground state $(n=1)$ to the fourth excited state $(n=4)$.The energy required is $E_{4} - E_{1} = \left(\frac{-E}{4^2}ight) - \left(\frac{-E}{1^2}ight)$.$= \frac{-E}{16} + E = \frac{15E}{16}$.Substituting $E = 20 \ eV$,we get:$E_{3rd\ excitation} = \frac{15 	imes 20}{16} = \frac{300}{16} = \frac{75}{4} \ V$.Thus,the correct option is $D$.

If the first excitation potential of a hypothetical hydrogen-like atom is $15 \ V$,then the third excitation potential of the atom is

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