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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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 $= 1.66 \times 10^{-27} \text{ kg}$,$h = 6.6 \times 10^{-34} \text{ Js}$]

Similar Questions

If the uncertainty in momentum of a particle is twice the uncertainty in its position,what will be the uncertainty in the position of the particle?

The uncertainty principle was enunciated by

The uncertainty of velocity of an electron is $5.7 \times 10^5 \ m \ s^{-1}$. Find its uncertainty in position.

Which of the following statements is incorrect?

Vedclass Test Series

Exam Paper Generator

Online Exam Module