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)

$\frac{1}{2}mV^{2} =\frac{k\left(Zq_{1}\right)q_{2}}{r}$
$\Rightarrow \frac{q_{2}}{m} =\frac{r .v^{2}}{2k q_{1}Z}$
$ \frac{q_{2}}{m}=\frac{2.5\times10^{-14} \times\left(2.1 \times10^{7}\right)^{^2}}{2\times9\times10^{9}\times79\times1.6\times10^{-19}} $
$\Rightarrow 4.84 \times10^{7} C/g$