In Bohr's atomic model,the electron is assume | vedclass

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(A) The time period $T$ for one revolution is given by $T = \frac{2 \pi r}{v}$.Substituting the values: $T = \frac{2 	imes (22/7) 	imes 0.5 	imes 1

Vedclass · Accepted Answer

$112$

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(A) The time period $T$ for one revolution is given by $T = \frac{2 \pi r}{v}$.Substituting the values: $T = \frac{2 	imes (22/7) 	imes 0.5 	imes 10^{-10}}{2.2 	imes 10^{6}}$.$T = \frac{2 	imes 22 	imes 0.5 	imes 10^{-10}}{7 	imes 2.2 	imes 10^{6}} = \frac{22 	imes 10^{-10}}{7 	imes 2.2 	imes 10^{6}} = \frac{10}{7} 	imes 10^{-16} \; s$.The current $I$ is given by $I = \frac{q}{T}$,where $q = e = 1.6 	imes 10^{-19} \; C$.$I = \frac{1.6 	imes 10^{-19}}{(10/7) 	imes 10^{-16}} = \frac{1.6 	imes 7}{10} 	imes 10^{-3} \; A$.$I = 1.12 	imes 10^{-3} \; A = 1.12 \; mA$.To express this as $.... 	imes 10^{-2} \; mA$,we write $1.12 \; mA = 112 	imes 10^{-2} \; mA$.Thus,the value is $112$.

In Bohr's atomic model,the electron is assumed to revolve in a circular orbit of radius $0.5 \times 10^{-10} \; m$. If the speed of the electron is $2.2 \times 10^{6} \; m/s$,then the current associated with the electron will be $.... \times 10^{-2} \; mA$. [Take $\pi$ as $\frac{22}{7}$]

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