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(C) The excitation potential is defined as the excitation energy divided by the elementary charge $e$.The minimum excitation energy corresponds to the

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$10.2$

Vedclass · Answer

(C) The excitation potential is defined as the excitation energy divided by the elementary charge $e$.The minimum excitation energy corresponds to the transition of an electron from the ground state $(n = 1)$ to the first excited state $(n = 2)$.The energy of the $n$-th orbit in a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \, eV$.For $n = 1$,$E_1 = -13.6 \, eV$.For $n = 2$,$E_2 = -\frac{13.6}{4} = -3.4 \, eV$.The minimum excitation energy is $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \, eV$.Therefore,the minimum excitation potential is $10.2 \, V$.

Minimum excitation potential of Bohr's first orbit in hydrogen atom is.....$V$

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