The energy of an electron in an orbit of a hy | vedclass

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(C) The radius of an orbit in a hydrogen-like ion is given by $r_n = a_0 	imes \frac{n^2}{Z}$, where $a_0 = 52.9 \ pm$ is the Bohr radius.Given $r_n

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$-8.72 	imes 10^{-18}$

Vedclass · Answer

(C) The radius of an orbit in a hydrogen-like ion is given by $r_n = a_0 	imes \frac{n^2}{Z}$, where $a_0 = 52.9 \ pm$ is the Bohr radius.Given $r_n = 52.9 \ pm$, we have $52.9 = 52.9 	imes \frac{n^2}{Z}$, which implies $\frac{n^2}{Z} = 1$ or $n^2 = Z$.The energy of an electron in a hydrogen-like ion is $E_n = E_1 	imes \frac{Z^2}{n^2}$, where $E_1 = -2.18 	imes 10^{-18} \ J$.Substituting $Z = n^2$ into the energy equation: $E_n = E_1 	imes \frac{(n^2)^2}{n^2} = E_1 	imes n^2$.For the first excited state $(n = 2)$, $E_2 = -2.18 	imes 10^{-18} 	imes 2^2 = -2.18 	imes 10^{-18} 	imes 4 = -8.72 	imes 10^{-18} \ J$.Thus, the correct option is $(C)$.

The energy of an electron in an orbit of a hydrogen-like ion with an orbit radius of $52.9 \ pm$ in $J$ is (ground state energy of an electron in a hydrogen atom is $-2.18 \times 10^{-18} \ J$).

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