Question
An alpha particle of energy $$5\,MeV$$ is scattered through $${180^ \circ }$$ by a fixed uranium nucleus. The distance of closest approach is of the order of
A.
$$1\, \mathop {\text{A}}\limits^ \circ $$
B.
$${10^{ - 10}}cm$$
C.
$${10^{ - 12}}cm$$
D.
$${10^{ - 15}}cm$$
Answer :
$${10^{ - 12}}cm$$
Solution :
One point charge is $$\left( {_{92}^{235}U} \right)$$ uranium nucleus
$$\therefore {q_1} = 92e$$
The other point charge is $$\alpha $$ particle $$\therefore {q_2} = + 2e$$
Here the loss in $$K.E.$$ = Gain in $$PE.$$ (till $$\alpha $$-particle reaches the distance $$d$$ )
$$\eqalign{ & \Rightarrow \frac{1}{2}m{v^2} = k\frac{{{q_1}{q_2}}}{r} \Rightarrow r = k\frac{{2{q_1}{q_2}}}{{\frac{1}{2}m{v^2}}} \cr & \therefore r = \frac{{9 \times {{10}^9} \times 2 \times 1.6 \times {{10}^{ - 19}} \times 92 \times 1.6 \times {{10}^{ - 19}}}}{{5 \times 1.6 \times {{10}^{ - 13}}}} \cr & = 529.92 \times {10^{ - 16}}m \cr & = 529.92 \times {10^{ - 14}}cm = 5.2992 \times {10^{ - 12}}cm \cr} $$
One point charge is $$\left( {_{92}^{235}U} \right)$$ uranium nucleus
$$\therefore {q_1} = 92e$$
The other point charge is $$\alpha $$ particle $$\therefore {q_2} = + 2e$$
Here the loss in $$K.E.$$ = Gain in $$PE.$$ (till $$\alpha $$-particle reaches the distance $$d$$ )
$$\eqalign{ & \Rightarrow \frac{1}{2}m{v^2} = k\frac{{{q_1}{q_2}}}{r} \Rightarrow r = k\frac{{2{q_1}{q_2}}}{{\frac{1}{2}m{v^2}}} \cr & \therefore r = \frac{{9 \times {{10}^9} \times 2 \times 1.6 \times {{10}^{ - 19}} \times 92 \times 1.6 \times {{10}^{ - 19}}}}{{5 \times 1.6 \times {{10}^{ - 13}}}} \cr & = 529.92 \times {10^{ - 16}}m \cr & = 529.92 \times {10^{ - 14}}cm = 5.2992 \times {10^{ - 12}}cm \cr} $$