Question

A beam of specific kind of particles of velocity
2.1 × 10⁷ m/s is scattered by a gold (Z = 79) nuclei. Find out specific charge (charge/mass) of this particle if the distance of closest approachis2.5 × 10^–14 m.

• A. 4.84 × 10⁷ C/g
• B. 4.84 × 10^–7 C/g
• C. 2.42 × 10⁷ C/g
• D. 3 × 10^–12 C/g

Answer 1

Answer: (A)

4.84 × 10⁷ C/g

Answer 2

$\frac{1}{2}$mv² =$\frac{k(Zq_{1})q_{2}}{r}$ Ž $\frac{q_{2}}{m}$=$\frac{r.v^{2}}{2k.q_{1}.Z}$

$\frac{q_{2}}{m}$=$\frac{2.5 \times 10^{- 14} \times (2.1 \times 10^{7})^{2}}{2 \times 9 \times 10^{9} \times 79 \times 1.6 \times 10^{- 19}}$Ž 4.84 × 10⁷ coulomb/g