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(C) The total energy $(E_n)$ of an electron in the $n^{th}$ shell is given by $E_n = \frac{E_1}{n^2}$,where $E_1$ is the energy of the ground state $(

Vedclass · Accepted Answer

$-800$

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(C) The total energy $(E_n)$ of an electron in the $n^{th}$ shell is given by $E_n = \frac{E_1}{n^2}$,where $E_1$ is the energy of the ground state $(n=1)$.Given $E_4 = -25 \, a.u.$,we have $-25 = \frac{E_1}{4^2}$.Therefore,$E_1 = -25 	imes 16 = -400 \, a.u.$For a hydrogen-like atom,the potential energy $(PE)$ is related to the total energy $(E)$ by the relation $PE = 2 	imes E$.Thus,for the ground state $(n=1)$,$PE_1 = 2 	imes E_1 = 2 	imes (-400) = -800 \, a.u.$

Supposing the energy of the fourth shell for a hydrogen atom is $-25 \, a.u.$ (arbitrary unit). What would be its potential energy in the ground state?

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