A proton is fired from very far away towards a nucleus with charge $Q=120 e$, where $e$ is the elect

Question

A proton is fired from very far away towards a nucleus with charge $Q=120 e$, where $e$ is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is (Take the proton mass, $m_p=\left(\frac{5}{3}ight) 	imes 10^{-27} \mathrm{~kg}$, \ $ \frac{h}{e}=4.2 	imes 10^{-15} \mathrm{~J} \mathrm{~s} / \mathrm{C}, \frac{1}{4 \pi \varepsilon_0} \ =9 	imes 10^9 \mathrm{~m} / \mathrm{F} $ $\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}ight)$

Professor PC Thomas Classes · Answer

(a) : As $\frac{1}{4 \pi \varepsilon_0} \frac{(120 e)(e)}{10 	imes 10^{-15}}=\frac{p^2}{2 m_p}$ \ where $p$ is the momentum of the proton and $m_p$ \ is the mass of the proton de Broglie wavelength of \ proton, $ \lambda=\frac{h}{p} \quad 	ext { or } \quad p=\frac{h}{\lambda} $ \ Substituting this value of $p$ in equation (i), we get \ $ \frac{1}{4 \pi \varepsilon_0} \frac{120 e^2}{10 	imes 10^{-15}}=\frac{h^2}{2 \lambda^2 m_p} ; \lambda^2=\frac{4 \pi \varepsilon_0 	imes 10 	imes 10^{-15} 	imes h^2}{2 m_p 	imes 120 e^2} $ \ Substituting the given numerical values, we get $ \ \begin{aligned} & \lambda^2=\frac{1 	imes 10 	imes 10^{-15} 	imes 4.2 	imes 10^{-15} 	imes 4.2 	imes 10^{-15}}{9 	imes 10^9 	imes 2 	imes \frac{5}{3} 	imes 10^{-27} 	imes 120} \ & =\frac{42 	imes 42 	imes 10^{-30} 	imes 10^{-14}}{36 	imes 10^{-14}}=49 	imes 10^{-30} \ & \lambda=7 	imes 10^{-15} \mathrm{~m}=7 \mathrm{fm} \end{aligned} $