The energy of an electron in the second Bohr | vedclass

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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = - \frac{2.178 	imes 10^{-18} \ J}{n^2}$.For th

Vedclass · Accepted Answer

$- 5.44 	imes 10^{-19} \ J$

Vedclass · Answer

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = - \frac{2.178 	imes 10^{-18} \ J}{n^2}$.For the second Bohr orbit,$n = 2$.Substituting the value of $n$ into the formula:$E_2 = - \frac{2.178 	imes 10^{-18}}{2^2} \ J$$E_2 = - \frac{2.178 	imes 10^{-18}}{4} \ J$$E_2 = - 5.445 	imes 10^{-19} \ J$.Thus,the energy is approximately $- 5.44 	imes 10^{-19} \ J$.

The energy of an electron in the second Bohr orbit of a hydrogen atom is:

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