The uncertainty in the position of an electro | vedclass

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(A) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.Given: $m = 9.1 	imes 10^{-28} \ g$,$v = 3.0 	ime

Vedclass · Accepted Answer

$1.92$

Vedclass · Answer

(A) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.Given: $m = 9.1 	imes 10^{-28} \ g$,$v = 3.0 	imes 10^4 \ cm \ s^{-1}$,and accuracy is $0.001\%$.First,calculate the uncertainty in velocity $(\Delta v)$:$\Delta v = v 	imes \frac{0.001}{100} = 3.0 	imes 10^4 	imes 10^{-5} = 0.3 \ cm \ s^{-1}$.Now,calculate the uncertainty in momentum $(\Delta p)$:$\Delta p = m 	imes \Delta v = 9.1 	imes 10^{-28} 	imes 0.3 = 2.73 	imes 10^{-28} \ g \ cm \ s^{-1}$.Using the uncertainty principle formula:$\Delta x = \frac{h}{4 \pi \cdot m \cdot \Delta v} = \frac{6.626 	imes 10^{-27}}{4 	imes 3.1416 	imes 2.73 	imes 10^{-28}}$.$\Delta x = \frac{6.626 	imes 10^{-27}}{34.287 	imes 10^{-28}} = \frac{66.26}{34.287} \approx 1.93 \ cm$.Rounding to the nearest provided option,the correct answer is $1.92 \ cm$.

The uncertainty in the position of an electron $(mass = 9.1 \times 10^{-28} \ g)$ moving with a velocity of $3.0 \times 10^4 \ cm \ s^{-1}$ accurate up to $0.001\%$ will be ................. $cm$ (Use $\frac{h}{4\pi}$ in the uncertainty expression,where $h = 6.626 \times 10^{-27} \ erg \ s$)

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