Question

A beam of specific kind of particles of velocity $2.1\times 10^7$ m/s is scattered by a gold $(Z=79)$ nuclei. Find out specific charge (charge/mass) of this particle if the distance of closest approach is $2.5\times 10^{-14}m$.

• A. $4.84\times 10^{7}C/g$
• B. $4.84\times 10^{-7}C/g$
• C. $2.42\times 10^{7}C/g$
• D. $3\times 10^{-12}C/g$

Answer 1

Answer: (A)

$\frac{1}{2}m{V}^2=\frac{k\left(Z{q}_1\right)q_2}{r}$
$\Rightarrow \frac{q_2}{m}=\frac{r.v^2}{2k{q}_{1}Z}$
$\frac{q_2}{m}=\frac{2.5\times 10^{-14}\times {\left(2.1\times 10^7\right)}^{{}^2}}{2\times 9\times 10^9\times 79\times 1.6\times 10^{-19}}$
$\Rightarrow 4.84\times 10^{7}C/g$