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 $10 \, fm$ to the nucleus. The de Broglie wavelength (in $fm$ ) of the proton at its start is: $\text{(}m_{\text{p}}=\left(5 / 3\right)\times 10^{- 2 7 \, }\text{kg}\text{;}\left( \textit{h}\right)/e=4\cdot 2\times 10^{- 1 5}\left(\text{ J s C}\right)^{- 1}\text{; }\frac{1}{4 \pi \left(\epsilon \right)_{0}}=9\times 10^{9}N \, \left(\text{m}\right)^{2}\left(\text{ C}\right)^{- 2}\text{; }1fm=10^{- 1 5} \, m)$

• A. $7$
• B. $5$
• C. $9$
• D. $3$

Answer 1

Answer: (A)

$0+\frac{1}{2} \mathrm{mv}^2=\frac{\mathrm{K}(\mathrm{Q}) \mathrm{e}}{10 \times 10^{-15}}=\frac{\mathrm{K}(120 \mathrm{e}) \mathrm{e}}{10 \times 10^{-15}}$
$\frac{1}{2} \times \frac{5}{3} \times 1 0^{- 2 7} \left(\text{v}\right)^{2} = \frac{9 \times 1 0^{9} \times 1 2 0 \times \left(1 \cdot 6 \times 1 0^{- 1 9}\right)^{2}}{1 0 \times 1 0^{- 1 5}}$
$\text{v} = \frac{9 \times 6 \times 1 0^{9} \times 1 2 0 \times 2 \cdot 5 6 \times 1 0^{- 3 8}}{5 0 \times 1 0^{- 4 2}}$
$\text{v} = \sqrt{3 3 1 \cdot 7 7 6 \times 1 0^{1 3}}$
$\lambda = \frac{\text{h}}{\text{mv}}$
$\lambda = \frac{4 \cdot 2 \times 1 0^{- 1 5} \times 1 \cdot 6 \times 1 0^{- 1 9}}{\frac{5}{3} \times 1 0^{- 2 7} \times \sqrt{3 3 1 \cdot 7 7 6 \times 1 0^{1 3}}} = \frac{4 \cdot 2 \times 4 \cdot 8 \times 1 0^{- 3 4}}{5 7 \cdot 6 \times 5 \times 1 0^{- 2 1}} = 0 \cdot 0 7 \times 1 0^{- 1 3}$
$\lambda = 7 \times 1 0^{- 1 5} = 7 \, \text{fm}$