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(D) The potential energy of an electron in the $n$-th orbit of a hydrogen atom is given by $V_n = -\frac{ke^2}{r_n} = -\frac{27.2}{n^2} 	ext{ eV}$.Th

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$5$

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(D) The potential energy of an electron in the $n$-th orbit of a hydrogen atom is given by $V_n = -\frac{ke^2}{r_n} = -\frac{27.2}{n^2} 	ext{ eV}$.Thus,the potential energy is inversely proportional to the square of the principal quantum number: $V \propto \frac{1}{n^2}$.Given the ratio $\frac{V_i}{V_f} = 6.25$,we can write:$\frac{V_i}{V_f} = \frac{n_f^2}{n_i^2} = 6.25$.Taking the square root on both sides:$\frac{n_f}{n_i} = \sqrt{6.25} = 2.5 = \frac{5}{2}$.This implies $n_f = 2.5 n_i$. Since $n_f$ and $n_i$ must be integers,we multiply by $2$ to get $n_f = 5$ and $n_i = 2$.Therefore,the smallest possible value for $n_f$ is $5$.

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number $n_i$ to another with quantum number $n_f$. $V_i$ and $V_f$ are respectively the initial and final potential energies of the electron. If $\frac{V_i}{V_f} = 6.25$,then the smallest possible $n_f$ is:

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