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 $10\,fm$ to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is (Take the proton mass, $m_{p}=(\frac{5}{3})\times10^{-27}\,kg$, $\frac{h}{e}=4.2\times10^{-15}J\, s/c$, $\frac{1}{4\pi\varepsilon_{0}}=9\times10^{9}\, m/F$, $1\,fm=10^{-15}\,m)$

• A. $7$
• B. $9$
• C. $11$
• D. $13$

Answer 1

Answer: (A)

As $\frac{1}{4\pi\varepsilon_{0}} \frac{\left(120e\right)\left(e\right)}{10\times10^{-15}}=\frac{p^{2}}{2m_{p}}\ldots\left(i\right)$
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
$\frac{1}{4\pi\varepsilon_{0}} \frac{120e^{2}}{10\times10^{-15}}=\frac{h^{2}}{2\lambda^{2} m_{p}}$
$\lambda^{2}=\frac{4\pi\varepsilon_{0}\times10\times10^{-15}\times h^{2}}{2m_{p}\times120e^{2}}$
Substituting the given numerical values, we get
$\lambda^{2}=\frac{1\times10\times10^{-15}\times4.2\times10^{-15}\times4.2\times10^{-15}}{9\times10^{9}\times2\times\frac{5}{3}\times10^{-27}\times120}$
$=\frac{42\times42\times10^{-30}\times10^{-14}}{36\times10^{-14}}=49\times10^{-30}$
$\lambda=7\times10^{-15}m=7\,fm$