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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = -\frac{2.18 	imes 10^{-18} 	ext{ J}}{n^2}$.Fo

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$-5.45 	imes 10^{-19} 	ext{ J}$

Vedclass · Answer

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = -\frac{2.18 	imes 10^{-18} 	ext{ J}}{n^2}$.For the stationary state corresponding to $n = 2$, we substitute the value of $n$ into the formula:$E_2 = -\frac{2.18 	imes 10^{-18} 	ext{ J}}{(2)^2}$$E_2 = -\frac{2.18 	imes 10^{-18} 	ext{ J}}{4}$$E_2 = -0.545 	imes 10^{-18} 	ext{ J}$$E_2 = -5.45 	imes 10^{-19} 	ext{ J}$.Therefore, the energy of the electron in the $n = 2$ state is $-5.45 	imes 10^{-19} 	ext{ J}$.

What is the energy of an electron in a hydrogen atom in a stationary state corresponding to $n = 2$ ?

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