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(A) At the point of closest approach,the initial kinetic energy $K$ of the proton is entirely converted into electrostatic potential energy $U$.$K = U

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) At the point of closest approach,the initial kinetic energy $K$ of the proton is entirely converted into electrostatic potential energy $U$.$K = U = \frac{1}{4 \pi \varepsilon_0} \frac{Q \cdot e}{r_0}$Given $Q = 120 \ e$ and $r_0 = 10 \ fm = 10 	imes 10^{-15} \ m$.$K = \frac{(9 	imes 10^9) 	imes (120 \ e) 	imes e}{10 	imes 10^{-15}} = \frac{p^2}{2m_0}$$p^2 = 2 m_0 	imes \frac{9 	imes 10^9 	imes 120 \ e^2}{10 	imes 10^{-15}} = 2 	imes \left(\frac{5}{3} 	imes 10^{-27}ight) 	imes 9 	imes 10^9 	imes 120 	imes 10^{15} 	imes e^2$$p^2 = 2 	imes \frac{5}{3} 	imes 9 	imes 120 	imes 10^{-27+9+15} 	imes e^2 = 3600 	imes 10^{-3} 	imes e^2 = 3.6 	imes e^2$Since $\lambda = \frac{h}{p}$,we have $p = \frac{h}{\lambda}$,so $p^2 = \frac{h^2}{\lambda^2}$.$\frac{h^2}{\lambda^2} = 3.6 	imes e^2 \implies \lambda^2 = \frac{h^2}{3.6 	imes e^2} = \frac{(h/e)^2}{3.6}$Given $h/e = 4.2 	imes 10^{-15} \ J \cdot s/C$.$\lambda^2 = \frac{(4.2 	imes 10^{-15})^2}{3.6} = \frac{17.64 	imes 10^{-30}}{3.6} = 4.9 	imes 10^{-30} \ m^2$$\lambda = \sqrt{49 	imes 10^{-31}} \approx 7 	imes 10^{-15} \ m = 7 \ fm$.

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