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(C) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \, eV$.For the ground state $(n = 1)$,$E_1 =

Vedclass · Accepted Answer

$1.632 	imes 10^{-18} \, J$

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(C) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \, eV$.For the ground state $(n = 1)$,$E_1 = -13.6 \, eV$.For the first excited state $(n = 2)$,$E_2 = -\frac{13.6}{2^2} = -3.4 \, eV$.The energy required to excite the electron from $n = 1$ to $n = 2$ is $\Delta E = E_2 - E_1 = -3.4 \, eV - (-13.6 \, eV) = 10.2 \, eV$.To convert this energy into Joules,we multiply by the conversion factor $1.6 	imes 10^{-19} \, J/eV$:$\Delta E = 10.2 	imes 1.6 	imes 10^{-19} \, J = 16.32 	imes 10^{-19} \, J = 1.632 	imes 10^{-18} \, J$.Thus,the correct option is $C$.

The energy required to excite an electron from the ground state of a hydrogen atom to the first excited state is:

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