From the following energy levels of a hydroge | vedclass

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Question

(C) The energy of an electron in the $n$th orbital of a hydrogen atom is given by the formula: $E_n = -2.18 	imes 10^{-18} \cdot \frac{Z^2}{n^2} \ J$

Vedclass · Accepted Answer

$0, -0.242 	imes 10^{-18}$

Vedclass · Answer

(C) The energy of an electron in the $n$th orbital of a hydrogen atom is given by the formula: $E_n = -2.18 	imes 10^{-18} \cdot \frac{Z^2}{n^2} \ J$. For a hydrogen atom, $Z = 1$. Thus, $E_n = -\frac{2.18 	imes 10^{-18}}{n^2} \ J$. For $n = \infty$, $E_{\infty} = -\frac{2.18 	imes 10^{-18}}{\infty} = 0 \ J$. For $n = 3$, $E_3 = -\frac{2.18 	imes 10^{-18}}{3^2} = -\frac{2.18 	imes 10^{-18}}{9} = -0.242 	imes 10^{-18} \ J$. Therefore, $E_{\infty} = 0 \ J$ and $E_3 = -0.242 	imes 10^{-18} \ J$. Hence, option $(c)$ is the correct answer.

From the following energy levels of a hydrogen atom, the values of $E_{\infty}$ and $E_3$ in $J$ are, respectively: $E_{\infty} = \ldots \ldots \ldots$ $E_3 = \ldots \ldots \ldots$ Given: $E_2 = -0.545 \times 10^{-18} \ J$, $E_1 = -2.18 \times 10^{-18} \ J$.

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