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(A) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$.Substituting $\Delta p = m \Delta v$,we get: $m \D

Vedclass · Accepted Answer

$22$

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(A) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$.Substituting $\Delta p = m \Delta v$,we get: $m \Delta v \Delta x = \frac{h}{4 \pi}$.Given: $\Delta v = 2.4 	imes 10^{-26} \, m \, s^{-1}$,$\Delta x = 10^{-7} \, m$,and $h = 6.626 	imes 10^{-34} \, J \, s$.$m 	imes (2.4 	imes 10^{-26}) 	imes (10^{-7}) = \frac{6.626 	imes 10^{-34}}{4 	imes 3.14159}$.$m 	imes 2.4 	imes 10^{-33} = \frac{6.626 	imes 10^{-34}}{12.566}$.$m = \frac{6.626 	imes 10^{-34}}{12.566 	imes 2.4 	imes 10^{-33}} = \frac{6.626}{30.1584} 	imes 10^{-1} \, kg$.$m \approx 0.2197 	imes 10^{-1} \, kg = 0.02197 \, kg$.Converting to grams: $m = 0.02197 	imes 1000 \, g = 21.97 \, g$.The nearest integer is $22$.

If the uncertainty in velocity and position of a minute particle in space are $2.4 \times 10^{-26} \, m \, s^{-1}$ and $10^{-7} \, m$ respectively,the mass of the particle in $g$ is $....$ (Nearest integer).
(Given: $h = 6.626 \times 10^{-34} \, J \, s$)

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If the uncertainty in velocity and position of a minute particle in space are $2.4 \times 10^{-26} \, m \, s^{-1}$ and $10^{-7} \, m$ respectively,the mass of the particle in $g$ is $....$ (Nearest integer).(Given: $h = 6.626 \times 10^{-34} \, J \, s$)