Question

In a Geiger-Marsden experiment. Find the distance of closest approach to the nucleus of a 7.7 MeV $\alpha -$before it comes momentarily to rest and reverses its direction. (Z for gold nucleus = 79)

• A. 10 fm
• B. 20 fm
• C. 30 fm
• D. 40 fm

Answer 1

Answer: (C)

30 fm

Answer 2

Led d be the distance of closest approach then by the conservation of energy

Initial kinetic energy of incoming $\alpha$particle, k = Final electric potential energy U of the system

As $K = \frac{1}{4\pi\varepsilon_{0}}\frac{(2e)(Ze)}{d}$

$\therefore d = \frac{1}{4\pi\varepsilon_{0}}\frac{2Ze^{2}}{K}$ ….. (i)

Here ,

$\frac{1}{4\pi\varepsilon_{0}} = 9 \times 10^{9}Nm^{2}C^{- 2},Z = 79,e = 1.6 \times 10^{- 19}C$

$K = 7.7MeV = 7.7 \times 10^{6} \times 1.6 \times 10^{- 19}J = 1.2 \times 10^{- 12}J$substituting these values in (i)

$d = \frac{2 \times 9 \times 10^{9} \times (1.6 \times 10^{- 19})^{2} \times 79}{1.2 \times 10^{- 12}}$

$d = 3 \times 10^{- 14}m$

$= 30m$ $(\because 1fm = 10^{- 15}m)$