Using Bohr's atomic model, the orbital period | vedclass

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(A) According to Bohr's model, the radius of the $n^{th}$ orbit is given by $r_n = \frac{n^2h^2\epsilon_0}{\pi me^2}$.The velocity of the electron in

Vedclass · Accepted Answer

$\frac{4\epsilon_0^2n^3h^3}{me^4}$

Vedclass · Answer

(A) According to Bohr's model, the radius of the $n^{th}$ orbit is given by $r_n = \frac{n^2h^2\epsilon_0}{\pi me^2}$.The velocity of the electron in the $n^{th}$ orbit is given by $v_n = \frac{e^2}{2\epsilon_0nh}$.The orbital period $T$ is the time taken to complete one revolution, which is $T = \frac{2\pi r_n}{v_n}$.Substituting the expressions for $r_n$ and $v_n$:$T = \frac{2\pi (\frac{n^2h^2\epsilon_0}{\pi me^2})}{(\frac{e^2}{2\epsilon_0nh})}$$T = \frac{2n^2h^2\epsilon_0}{me^2} 	imes \frac{2\epsilon_0nh}{e^2}$$T = \frac{4\epsilon_0^2n^3h^3}{me^4}$.Thus, the correct option is $A$.

Using Bohr's atomic model, the orbital period of an electron in a hydrogen atom in the $n^{th}$ orbit is ($\epsilon_0$ = permittivity of free space, $h$ = Planck's constant, $m$ = mass of electron, $e$ = electronic charge).

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