Calculate the mass of an elementary particle, | vedclass

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(D) Given: Precision $= \pm 1 \%$,$\Delta x = 1.05 	imes 10^{-13} \ m$,$h = 6.6 	imes 10^{-34} \ kg \ m^2 \ s^{-1}$.Velocity of light $c = 3 	imes

Vedclass · Accepted Answer

$0.8 	imes 10^{-28} \ kg$

Vedclass · Answer

(D) Given: Precision $= \pm 1 \%$,$\Delta x = 1.05 	imes 10^{-13} \ m$,$h = 6.6 	imes 10^{-34} \ kg \ m^2 \ s^{-1}$.Velocity of light $c = 3 	imes 10^8 \ m/s$.Particle velocity $v = 2c = 6 	imes 10^8 \ m/s$.Uncertainty in velocity $\Delta v = 1 \% 	ext{ of } v = 0.01 	imes 6 	imes 10^8 = 6 	imes 10^6 \ m/s$.According to Heisenberg's Uncertainty Principle: $\Delta x \cdot m \cdot \Delta v \ge \frac{h}{4 \pi}$.$m = \frac{h}{4 \pi \cdot \Delta x \cdot \Delta v}$.$m = \frac{6.6 	imes 10^{-34}}{4 	imes 3.14 	imes 1.05 	imes 10^{-13} 	imes 6 	imes 10^6}$.$m = \frac{6.6 	imes 10^{-34}}{79.128 	imes 10^{-7}} \approx 0.0834 	imes 10^{-27} \ kg = 0.834 	imes 10^{-28} \ kg$.Rounding to the nearest provided option,the correct value is $0.8 	imes 10^{-28} \ kg$.

Calculate the mass of an elementary particle,which is accelerated to twice the velocity of light with the precision $\pm 1 \%$,and has $1.05 \times 10^{-13} \ m$ uncertainty in position. $(h = 6.6 \times 10^{-34} \ kg \ m^2 \ s^{-1})$

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