Question

A proton is fired from far away toward the nucleus of a mercury atom. Mercury is element number $80$, and the diameter of the nucleus is$14.0fm$ . If the proton is fired at a speed of$4.0\times {10}^{7}m/s$, what is its closest approach to the surface of the nucleus? Assume the nucleus remains at rest

Short answer

The proton's closest approach to the surface of the nucleus is$\mathbf{6}\mathbf{.}\mathbf{8}\mathbf{}\mathit{f}\mathit{m}$

Step-by-step solution

Given information and theory used 

Given :

Mercury is element number : 80

Diameter of the nucleus is : 14.0 fm.

The proton is fired at a speed of : $4.0\times {10}^{7}m/s$

Theory used :

From the equation of conservation of energy, we have :

$\frac{1}{2}m{v}^2=q∆V$

Calculating the proton's closest approach to the surface of the nucleus

We can treat the nucleus' charge as a point charge and get

$\frac{1}{2}m{v}^2=k\frac{80e^2}{d/2}$

from energy conservation, where $d$is the distance from the point charge — that is, the middle of the nucleus to the point where the proton stops.

This distance can be calculated as $d=\frac{320e^{2}k}{m{v}^2}$,

where $m$is the proton's mass.

This means that the distance a from the nucleus' surface is $a=\frac{d-D}{2}$, where $D$is the nucleus' diameter.

As a result, we may say that the value will be

$a=\frac{160e^{2}k}{m{v}^2}-\frac{D}{2}$

In terms of numbers, we have

$$\begin{gathered} a=\frac{160\times {(1.6\times {10}^{-19})}^2\times 9\times 10⁹}{1.67\times {10}^{-27}·{(4\times {10}^7)}^2}-\frac{14\times {10}^{-15}}{2} \\ =\boxed{\mathbf{6}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{}\mathit{m}\mathbf{}} \end{gathered}$$

The result can be written as $\mathit{a}\mathbf{}\mathbf{=}\mathbf{}\mathbf{6}\mathbf{.}\mathbf{8}\mathbf{}\mathit{f}\mathit{m}$.