If a proton is accelerated to a velocity of 3 | vedclass

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(D) The uncertainty in velocity $\Delta v$ is given by $0.5 \%$ of $3 	imes 10^7 	ext{ ms}^{-1}$.$\Delta v = \frac{0.5}{100} 	imes 3 	imes 10^7

Vedclass · Accepted Answer

$2.11 	imes 10^{-13} 	ext{ m}$

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(D) The uncertainty in velocity $\Delta v$ is given by $0.5 \%$ of $3 	imes 10^7 	ext{ ms}^{-1}$.$\Delta v = \frac{0.5}{100} 	imes 3 	imes 10^7 	ext{ ms}^{-1} = 1.5 	imes 10^5 	ext{ ms}^{-1}$.According to Heisenberg's uncertainty principle,$\Delta x 	imes \Delta p \ge \frac{h}{4 \pi}$.Since $\Delta p = m \Delta v$,we have $\Delta x = \frac{h}{4 \pi m \Delta v}$.Substituting the values: $\Delta x = \frac{6.6 	imes 10^{-34}}{4 	imes 3.14 	imes 1.66 	imes 10^{-27} 	imes 1.5 	imes 10^5}$.$\Delta x = \frac{6.6 	imes 10^{-34}}{31.27 	imes 10^{-22}} \approx 2.11 	imes 10^{-13} 	ext{ m}$.

If a proton is accelerated to a velocity of $3 \times 10^7 \text{ ms}^{-1}$ which is accurate up to $\pm 0.5 \%$,then the uncertainty in its position will be $\ldots \ldots \ldots$ [mass of proton $= 1.66 \times 10^{-27} \text{ kg}$,$h = 6.6 \times 10^{-34} \text{ Js}$]

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