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(B) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta v \geq \frac{h}{4 \pi m}$.Given: $\Delta x = 10.98 \ nm = 10.98 	imes 10^{

Vedclass · Accepted Answer

$\frac{1.6565 	imes 10^4}{\pi}$

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(B) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta v \geq \frac{h}{4 \pi m}$.Given: $\Delta x = 10.98 \ nm = 10.98 	imes 10^{-9} \ m$,$m = 9.11 	imes 10^{-31} \ kg$,$h = 6.63 	imes 10^{-34} \ Js$.Substituting the values:$\Delta v = \frac{h}{4 \pi m \Delta x} = \frac{6.63 	imes 10^{-34}}{4 	imes \pi 	imes 9.11 	imes 10^{-31} 	imes 10.98 	imes 10^{-9}}$.$\Delta v = \frac{6.63 	imes 10^{-34}}{4 	imes \pi 	imes 10.00278 	imes 10^{-39}}$.$\Delta v = \frac{6.63 	imes 10^5}{4 	imes \pi 	imes 10.00278} = \frac{1.6565 	imes 10^4}{\pi} \ ms^{-1}$.

If a suitable photon is employed to locate an electron (mass $= 9.11 \times 10^{-31} \ kg$) in an atom within a distance of $10.98 \ nm$,the uncertainty involved in the measurement of its velocity in $ms^{-1}$ is

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