What is the potential energy of the electron | vedclass

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(B) The principal quantum number for the $L$ shell is $n = 2$. The total energy of an electron in the $n^{th}$ Bohr orbit is given by $E_n = -\frac{13

Vedclass · Accepted Answer

$-6.8$

Vedclass · Answer

(B) The principal quantum number for the $L$ shell is $n = 2$. The total energy of an electron in the $n^{th}$ Bohr orbit is given by $E_n = -\frac{13.6 \ eV}{n^2}$. The potential energy $(P.E.)$ of an electron is related to the total energy by the relation $P.E. = 2 	imes E_n$. Substituting the expression for $E_n$: $P.E. = 2 	imes (-\frac{13.6 \ eV}{n^2}) = -\frac{27.2 \ eV}{n^2}$. For the $L$ shell $(n = 2)$: $P.E. = -\frac{27.2 \ eV}{2^2} = -\frac{27.2 \ eV}{4} = -6.8 \ eV$.

What is the potential energy of the electron in the $L$ shell of the hydrogen atom? (in $eV$)

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