Question

A proton is fired from very far away towards a nucleus with charge $Q=120e$, where $e$ is the electronic charge. It makes a closest approach of $10fm$ to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is (Take the proton mass, $m_p=(\frac{5}{3})\times 10^{-27}kg$, $\frac{h}{e}=4.2\times 10^{-15}Js/c$, $\frac{1}{4\pi {\epsilon }_0}=9\times 10^{9}m/F$, $1fm=10^{-15}m)$

• A. $7$
• B. $9$
• C. $11$
• D. $13$

Answer 1

Answer: (A)

As $\frac{1}{4\pi {\epsilon }_0}\frac{\left(120e\right)\left(e\right)}{10\times 10^{-15}}=\frac{p^2}{2m_p}\dots \left(i\right)$
where $p$ is the momentum of the proton and $m_p$ is the mass of the proton
de Broglie wavelength of proton,
$\lambda =\frac{h}{p}$ or $p=\frac{h}{\lambda }$
Substituting this value of $p$ in equation $\left(i\right)$, we get
$\frac{1}{4\pi {\epsilon }_0}\frac{120e^2}{10\times 10^{-15}}=\frac{h^2}{2{\lambda }^2{m}_p}$
${\lambda }^2=\frac{4\pi {\epsilon }_0\times 10\times 10^{-15}\times h^2}{2m_p\times 120e^2}$
Substituting the given numerical values, we get
${\lambda }^2=\frac{1\times 10\times 10^{-15}\times 4.2\times 10^{-15}\times 4.2\times 10^{-15}}{9\times 10^9\times 2\times \frac{5}{3}\times 10^{-27}\times 120}$
$=\frac{42\times 42\times 10^{-30}\times 10^{-14}}{36\times 10^{-14}}=49\times 10^{-30}$
$\lambda =7\times 10^{-15}m=7fm$