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(D) Given: Radius $r = 0.53 	ext{ Å} = 0.53 	imes 10^{-10} 	ext{ m}$.Time period $T = 1.571 	imes 10^{-16} 	ext{ s}$.The velocity $v$ of the elec

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$2.12 	imes 10^{6} 	ext{ m/s}$

Vedclass · Answer

(D) Given: Radius $r = 0.53 	ext{ Å} = 0.53 	imes 10^{-10} 	ext{ m}$.Time period $T = 1.571 	imes 10^{-16} 	ext{ s}$.The velocity $v$ of the electron is given by the formula $v = \frac{	ext{Distance}}{	ext{Time}} = \frac{2 \pi r}{T}$.Substituting the values: $v = \frac{2 	imes 3.142 	imes 0.53 	imes 10^{-10}}{1.571 	imes 10^{-16}}$.Since $2 	imes 3.142 = 6.284$,we have $v = \frac{6.284 	imes 0.53 	imes 10^{-10}}{1.571 	imes 10^{-16}}$.Note that $\frac{6.284}{1.571} = 4$.So,$v = 4 	imes 0.53 	imes 10^{6} 	ext{ m/s}$.$v = 2.12 	imes 10^{6} 	ext{ m/s}$.

The electron in a hydrogen atom is moving in an orbit of radius $0.53 \text{ Å}$. It takes $1.571 \times 10^{-16} \text{ s}$ to complete one revolution. The velocity of the electron will be $[\pi = 3.142]$.

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