Question

The distance of the closest approach when a $5.0\text{MeV}$ proton approaches a gold nucleus is $(Z=79)$

• A. $47\text{fermi}$
• B. $39\text{fermi}$
• C. $23\text{fermi}$
• D. $16\text{fermi}$

Answer 1

The correct option is C $23\text{fermi}$

By conservation of energy,

$K.E$ of the proton = Electrostatic $P.E$ of proton and target nucleus at closest distance.

$\Rightarrow K.E=\frac{1}{4\pi {\epsilon }_0}\frac{\left(Ze\right)\left(e\right)}{r}$

$\Rightarrow r=\frac{1}{4\pi {\epsilon }_0}\frac{Z{e}^2}{K.E}---\left(1\right)$

Given that,
$K.E=5.0\text{MeV}=5\times 1.6\times {10}^{-13}\text{J}$, $Z=79$

We know that,

$e=1.6\times {10}^{-19}\text{C};\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9$

Substituting the above values in equation $\left(1\right)$, we get,

$r=\frac{9\times {10}^9\times 79\times (1.6\times {10}^{-19}{)}^2}{5\times 1.6\times {10}^{-13}}$

$r=\frac{9\times 79\times 1.6\times {10}^{-38}\times {10}^{22}}{5}$

$r=227.52\times {10}^{-16}\text{m}$

$\Rightarrow r\approx 23\times {10}^{-15}m$

So, The distance of the closest approach,

$r=23\text{fm}$

Hence, option $\left(C\right)$ is correct.