In the n^{th} orbit,the energy of an electron | vedclass

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(A) The energy of an electron in the $n^{th}$ orbit is $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.For the first orbit $(n=1)$: $E_1 = -\frac{13.6}{1^2} = -1

Vedclass · Accepted Answer

$10.2$

Vedclass · Answer

(A) The energy of an electron in the $n^{th}$ orbit is $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.For the first orbit $(n=1)$: $E_1 = -\frac{13.6}{1^2} = -13.6 	ext{ eV}$.For the second orbit $(n=2)$: $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 	ext{ eV}$.The energy required to excite the electron from the first orbit to the second orbit is given by the difference in energy levels:$\Delta E = E_2 - E_1$$\Delta E = -3.4 	ext{ eV} - (-13.6 	ext{ eV})$$\Delta E = -3.4 + 13.6 = 10.2 	ext{ eV}$.Therefore,the energy required is $10.2 	ext{ eV}$.

In the $n^{th}$ orbit,the energy of an electron is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$ for a hydrogen atom. The energy required to excite the electron from the first orbit $(n=1)$ to the second orbit $(n=2)$ will be..... eV.

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