Chapter 30, Problem 6P

To determine

The acceleration of an alpha particle subjected to the electrostatic force.

Answer to Problem 6P

Solution:

$a=9.6\times {10}^{27}m/s^2$

Explanation of Solution

Electrostatic force of repulsion is given by,

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{{(2r)}^2}$

Where q is the charge on alpha particle, q=2× $1.602\times {10}^{-19}C$

r is radius of alpha particle, $r=r_0\sqrt[3]{A}$

$\begin{array}{l}r=r_0\sqrt[3]{A}\\ r=1.25\times \sqrt[3]{4}\\ r=1.25\times 1.587\\ r=1.9\text{fermi}\\ \therefore \text{2}r\text{=3}\text{.8fermi}\end{array}$

$\begin{array}{l}F=9\times {10}^9\frac{{(3.2\times {10}^{-19})}^2}{{(3.8\times {10}^{-15})}^2}\\ F=63.8\text{N}\end{array}$

According to Newton’s second law of motion,

$\begin{array}{l}F=ma\\ a=\frac{F}{m}\\ m=M_{He}-M_e\\ m=(4.002)(931.5)MeV/c^2-2(0.511)MeV/c^2\\ m=3726.841MeV/c^2\\ a=\frac{63.8}{3726.84\times \frac{1.66\times {10}^{-27}}{931.5}}\\ a=9.6\times {10}^{27}m/s^2\end{array}$