Question

A proton is fired from very far away towards a nucleus with charge $Q=120e$ , where $e$ is the electronic charge. It makes the closest approach of $10fm$ to the nucleus. The de Broglie wavelength (in $fm$ ) of the proton at its start is: $\text{(}{m}_{\text{p}}=\left(5/3\right)\times 10^{-27}\text{kg}\text{;}\left(\text{h}\right)/e=4\cdot 2\times 10^{-15}{\left(\text{ J s C}\right)}^{-1}\text{; }\frac{1}{4\pi {\left(ϵ\right)}_0}=9\times 10^{9}N{\left(\text{m}\right)}^2{\left(\text{ C}\right)}^{-2}\text{; }1fm=10^{-15}m)$

• A. $7$
• B. $5$
• C. $9$
• D. $3$

Answer 1

Answer: (A)

$0+\frac{1}{2}{\mathrm{mv}}^2=\frac{\mathrm{K}(\mathrm{Q})\mathrm{e}}{10\times 10^{-15}}=\frac{\mathrm{K}(120\mathrm{e})\mathrm{e}}{10\times 10^{-15}}$
$\frac{1}{2}\times \frac{5}{3}\times 10^{-27}{\left(\text{v}\right)}^2=\frac{9\times 10^9\times 120\times {\left(1\cdot 6\times 10^{-19}\right)}^2}{10\times 10^{-15}}$
$\text{v}=\frac{9\times 6\times 10^9\times 120\times 2\cdot 56\times 10^{-38}}{50\times 10^{-42}}$
$\text{v}=\sqrt{331\cdot 776\times 10^{13}}$
$\lambda =\frac{\text{h}}{\text{mv}}$
$\lambda =\frac{4\cdot 2\times 10^{-15}\times 1\cdot 6\times 10^{-19}}{\frac{5}{3}\times 10^{-27}\times \sqrt{331\cdot 776\times 10^{13}}}=\frac{4\cdot 2\times 4\cdot 8\times 10^{-34}}{57\cdot 6\times 5\times 10^{-21}}=0\cdot 07\times 10^{-13}$
$\lambda =7\times 10^{-15}=7\text{fm}$