According to de-Broglie, the de-Broglie wavel | vedclass

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(C) According to the Bohr quantization condition, the circumference of the orbit is an integral multiple of the de-Broglie wavelength: $2 \pi r_n = n

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) According to the Bohr quantization condition, the circumference of the orbit is an integral multiple of the de-Broglie wavelength: $2 \pi r_n = n \lambda$.Given: $\lambda = 10^{-9} \, m$.The radius of the $n^{th}$ orbit of a hydrogen atom is given by $r_n = n^2 a_0$, where $a_0 \approx 0.529 	imes 10^{-10} \, m$.Substituting $r_n$ into the condition: $2 \pi (n^2 a_0) = n \lambda$.Simplifying for $n$: $n = \frac{\lambda}{2 \pi a_0}$.Substituting the values: $n = \frac{10^{-9}}{2 	imes 3.14159 	imes 0.529 	imes 10^{-10}} \approx \frac{10^{-9}}{3.32 	imes 10^{-10}} \approx 3.01$.Thus, the principal quantum number is $n = 3$.

According to de-Broglie, the de-Broglie wavelength for an electron in an orbit of a hydrogen atom is $10^{-9} \, m$. The principal quantum number for this electron is:

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