An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of

Question

An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of (A) 1 mathringA (B) 10-10 cm (C) 10-12 cm (D) 10-15 cm

Tardigrade Admin · Accepted Answer

From conservation of mechanical energy decrease in kinetic energy = increase in potential energy or

\frac{1}{4 π ε _{0}} \frac{( Z e ) ( Z e )}{r _{min}} = 5 M e V = 5 \times 1.6 \times 1 0^{- 13} J

∴ r_{min} = \frac{1}{4 π ε _{0}} \frac{2 Z e}{5 \times 1.6 \times 1 0 ^{- 13}}

= \frac{( 9 \times 1 0 ^{9} ) ( 2 ) ( 92 ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{5 \times 1.6 \times 1 0 ^{- 13}}

= 5.3 \times 1 0^{- 14} m = 5.3 \times 1 0^{- 12} c m

i.e.

r_{min}

is of the order of

1 0^{- 12} c m

Q. An alpha particle of energy $5 M e V$ is scattered through $18 0^{\circ}$ by a fixed uranium nucleus. The distance of closest approach is of the order of

$1 \overset{˚}{A}$

$1 0^{- 10} c m$

$1 0^{- 12} c m$

$1 0^{- 15} c m$

Q. An alpha particle of energy 5MeV is scattered through 180∘ by a fixed uranium nucleus. The distance of closest approach is of the order of

1A˚

10−10cm

10−12cm

10−15cm

Q. An alpha particle of energy $5 M e V$ is scattered through $18 0^{\circ}$ by a fixed uranium nucleus. The distance of closest approach is of the order of

$1 \overset{˚}{A}$

$1 0^{- 10} c m$

$1 0^{- 12} c m$

$1 0^{- 15} c m$