The number of revolutions per second made by | vedclass

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Question

(D) According to Bohr's quantization condition,the angular momentum $L$ is given by $mvr = \frac{h}{2\pi}$.Since $v = r\omega = r(2\pi
u)$,where $
u

Vedclass · Accepted Answer

$10^{15}$

Vedclass · Answer

(D) According to Bohr's quantization condition,the angular momentum $L$ is given by $mvr = \frac{h}{2\pi}$.Since $v = r\omega = r(2\pi
u)$,where $
u$ is the frequency of revolution:$m(r \cdot 2\pi
u)r = \frac{h}{2\pi}$$\Rightarrow 
u = \frac{h}{4\pi^2 mr^2}$Substituting the values $h = 6.6 	imes 10^{-34} \ J \cdot s$,$m = 9.1 	imes 10^{-31} \ kg$,and $r = 0.53 	imes 10^{-10} \ m$:$
u = \frac{6.6 	imes 10^{-34}}{4 	imes (3.14)^2 	imes 9.1 	imes 10^{-31} 	imes (0.53 	imes 10^{-10})^2}$$
u \approx 6.5 	imes 10^{15} \ 	ext{rev/sec}$.Thus,the order of magnitude is $10^{15}$.

The number of revolutions per second made by an electron in the first Bohr orbit of a hydrogen atom is of the order of

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