The -particles with initial kinetic energy 4-8 mev are shot at gold atoms z-79- The potential of -particle at the closest distance to the gold nucleus is

An alpha particle of $5 M e V$ at a large distance proceeds towards a gold nucleus $(Z = 79)$ to make a head on collision. The closest distance of approach from the centre of gold nucleus is:

Q. An

α -

particle of energy

5 M e V

is scattered through

180^{\circ}

by gold nucleus. The distance of closest approach is of the order of

Q. An

α - particle

with kinetic energy

10 MeV

is heading towards a stationary nucleus of atomic number

50

. Calculate the distance of closest approach. Initially they were far apart.

Join BYJU'S Learning Program

The α-particles with initial kinetic energy 4.8 mev are shot at gold atoms (z=79). The potential of α-particle at the closest distance to the gold nucleus is

The correct option is D 2.4MVThe energy of alpha particle is E=4.8MeVAlso, for alpha particles (e=2e)E=2eVV=E2eV=4.8×106×1.6×10−192(1.6×10−19)V=4.8×1062=2.4×106=2.4MV

The $α$ -particles with initial kinetic energy $4.8$ mev are shot at gold atoms $(z = 79)$ . The potential of $α$ -particle at the closest distance to the gold nucleus is