Angular momentum of an electron in a hydrogen | vedclass

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(B) According to Bohr's postulate,the angular momentum $L$ is given by $L = \frac{nh}{2\pi}$.Given $L = \frac{3h}{2\pi}$,we find that the principal qu

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$10$

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(B) According to Bohr's postulate,the angular momentum $L$ is given by $L = \frac{nh}{2\pi}$.Given $L = \frac{3h}{2\pi}$,we find that the principal quantum number $n = 3$.The radius of the $n$-th orbit is given by $r_n = n^2 r_0$,where $r_0 = 0.529 \mathring{A}$.For $n = 3$,$r_3 = 3^2 	imes 0.529 \mathring{A} = 9 	imes 0.529 \mathring{A} = 4.761 \mathring{A}$.According to the de Broglie hypothesis,the circumference of the orbit is equal to $n$ times the wavelength: $2\pi r_n = n\lambda$.Therefore,$\lambda = \frac{2\pi r_n}{n} = \frac{2\pi 	imes 9r_0}{3} = 6\pi r_0$.Substituting $r_0 = 0.529 \mathring{A}$,we get $\lambda = 6 	imes 3.1416 	imes 0.529 \mathring{A} \approx 9.97 \mathring{A} \approx 10 \mathring{A}$.

Angular momentum of an electron in a hydrogen atom is $\frac{3h}{2\pi}$. The wavelength of this electron is approximately $...... \mathring{A}$.

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