What is the angular momentum of an electron i | vedclass

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = -\frac{13.6}{n^2} \ eV$.Given $E_n = -0.544 \ e

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$\frac{5h}{2\pi}$

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = -\frac{13.6}{n^2} \ eV$.Given $E_n = -0.544 \ eV$,we set up the equation: $-0.544 = -\frac{13.6}{n^2}$.Solving for $n^2$: $n^2 = \frac{13.6}{0.544} = 25$.Thus,$n = 5$.According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by: $L = n\frac{h}{2\pi}$.Substituting $n = 5$,we get: $L = 5\frac{h}{2\pi} = \frac{5h}{2\pi}$.

What is the angular momentum of an electron in Bohr's hydrogen atom whose energy is $-0.544 \ eV$?

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