Question

A beam of some kind of particle of velocity $2\times {10}^{7}m/s$ is scattered by a gold $(z=79)$ foil. Find specific charge of this particle (charge/mass) if the distance of closest approach is $7.9\times {10}^{-14}m$.

• A. $4.5\times {10}^{-4}$
• B. $9\times {10}^3$
• C. $1.3\times {10}^8$
• D. $4.3\times {10}^8$

Answer 1

The correct option is C $1.3\times {10}^8$

From the formula

$\frac{1}{2}m{v}^2=\frac{k{q}_1{q}_2}{r}$

Where $m=$ Mass of particle

$r=$ Distance of closest approach

$v=$ velocity

$q_1=$ Charge on particle

$q_2=$ Charge on metaL foil nuclei

Given velocity$=2\times {10}^7$ m/s

$r=7.9\times {10}^{-14}$ m

$Z$ of gold $=79$

$\frac{1}{r}=\frac{m{v}^2}{2{kq}_1{q}_2}$

$\frac{1}{7.9\times {10}^{-14}}=\frac{m(2\times {10}^{-7}{)}^2}{2\times 9\times {10}^{-9}\times 79\times 1.6\times {10}^{-19}\times e}$

$\frac{e}{m}=1.3\times {10}^8$

Hence, the correct option is C.