An α -particle of 10 MeV collides head on with a copper nucleus (Z=29) and is deflected back. Then the minimum distance of approach between the centres of the two is

Question

An α -particle of 10 MeV collides head on with a copper nucleus (Z=29) and is deflected back. Then the minimum distance of approach between the centres of the two is (A) 8.4 × 10-15 cm (B) 8.4 × 10-15 m (C) 4.2 × 10-15 m (D) 4.2 × 10-15 cm

Tardigrade Admin · Accepted Answer

Let the minimum distance of approach be

r_{0}

. At this distance, the whole of the kinetic energy of the alpha-particle will be converted into the electrical potential energy.
The positive charge on the copper nucleus

= Z e = 29 e

and the positive charge on the

α

-particle

= 2 e

At

r = r_{0}, K E = PE

10 \times (1.6 \times 1 0^{- 13}) = \frac{1}{4 π ε _{0}} \frac{( 29 e ) ( 2 e )}{r _{0}}

∴ r_{0} = \frac{( 9 \times 1 0 ^{9} ) ( 29 ) ( 2 ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{10 \times ( 1.6 \times 1 0 ^{- 13} )} = 8.4 \times 1 0^{- 15} m

Q. An $α$ -particle of $10 M e V$ collides head on with a copper nucleus $(Z = 29)$ and is deflected back. Then the minimum distance of approach between the centres of the two is

$8.4 \times 1 0^{- 15} c m$

$8.4 \times 1 0^{- 15} m$

$4.2 \times 1 0^{- 15} m$

$4.2 \times 1 0^{- 15} c m$

Q. An α -particle of 10MeV collides head on with a copper nucleus (Z=29) and is deflected back. Then the minimum distance of approach between the centres of the two is

8.4×10−15cm

8.4×10−15m

4.2×10−15m

4.2×10−15cm

Q. An $α$ -particle of $10 M e V$ collides head on with a copper nucleus $(Z = 29)$ and is deflected back. Then the minimum distance of approach between the centres of the two is

$8.4 \times 1 0^{- 15} c m$

$8.4 \times 1 0^{- 15} m$

$4.2 \times 1 0^{- 15} m$

$4.2 \times 1 0^{- 15} c m$