A proton and an anti-proton come close to eac | vedclass

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Question

(A) The situation involves a proton and an anti-proton approaching each other. Since they have opposite charges,they attract each other.Let the veloci

Vedclass · Accepted Answer

$1.17$

Vedclass · Answer

(A) The situation involves a proton and an anti-proton approaching each other. Since they have opposite charges,they attract each other.Let the velocity of each particle be $v$ at a distance $r = 10 \, cm = 0.1 \, m$.From the law of conservation of energy,the total energy at infinity (where potential energy is zero and assuming they start from rest) is equal to the total energy at distance $r$.$(PE)_{i} + (KE)_{i} = (PE)_{f} + (KE)_{f}$$0 + 0 = -\frac{K e^2}{r} + \frac{1}{2} m v^2 + \frac{1}{2} m v^2$Note: The potential energy is negative because the charges are opposite.$\frac{K e^2}{r} = m v^2$$v = \sqrt{\frac{K e^2}{m r}}$Substituting the values: $K = 9 	imes 10^9 \, N m^2/C^2$,$e = 1.6 	imes 10^{-19} \, C$,$m = 1.67 	imes 10^{-27} \, kg$,$r = 0.1 \, m$.$v = \sqrt{\frac{9 	imes 10^9 	imes (1.6 	imes 10^{-19})^2}{1.67 	imes 10^{-27} 	imes 0.1}}$$v = \sqrt{\frac{9 	imes 2.56 	imes 10^{-29}}{1.67 	imes 10^{-28}}} = \sqrt{\frac{23.04 	imes 10^{-1}}{1.67}} = \sqrt{13.79} \approx 3.71 \, m/s$.Wait,re-evaluating the provided solution logic: The original solution provided in the prompt used $m = 1.67 	imes 10^{-27}$ and calculated $1.17 \, m/s$. Let's re-check the calculation: $\sqrt{\frac{9 	imes 10^9 	imes 2.56 	imes 10^{-38}}{1.67 	imes 10^{-27} 	imes 0.1}} = \sqrt{\frac{23.04 	imes 10^{-29}}{1.67 	imes 10^{-28}}} = \sqrt{13.79 	imes 0.1} = \sqrt{1.379} \approx 1.17 \, m/s$. The calculation is correct.

$A$ proton and an anti-proton come close to each other in vacuum such that the distance between them is $10 \, cm$. Consider the potential energy to be zero at infinity. The velocity at this distance will be ........... $\, m/s$.

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