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(D) The frequency of revolution $f$ of an electron in the $n^{th}$ orbit is given by $f = \frac{v_n}{2\pi r_n}$.For the first Bohr orbit $(n=1)$ of a

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$6.57 	imes 10^{15}$

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(D) The frequency of revolution $f$ of an electron in the $n^{th}$ orbit is given by $f = \frac{v_n}{2\pi r_n}$.For the first Bohr orbit $(n=1)$ of a hydrogen atom,the velocity $v_1 = 2.18 	imes 10^6 \, m/s$ and the radius $r_1 = 0.529 	imes 10^{-10} \, m$.Substituting these values:$f = \frac{2.18 	imes 10^6}{2 	imes 3.14159 	imes 0.529 	imes 10^{-10}}$$f \approx \frac{2.18 	imes 10^6}{3.324 	imes 10^{-10}}$$f \approx 6.56 	imes 10^{15} \, Hz$.Thus,the electron completes approximately $6.57 	imes 10^{15}$ revolutions per second.

How many times does the electron go round the first Bohr orbit in a second?

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