A 24.6 MeV alpha particle moves towards gold nucleus (atomic number 79). Calculate the distance of closest approach of alpha particle.

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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

3 MeV

is scattered through

180^{\circ}

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

The atomic number of gold is

79

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.

Q. 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

Q. The maximum K.E found in alpha particle of natural origin is

7.7

MeV. Find the closest distance of approach for the gold nucleus

(Z = 79

for gold

)

[Note : At closest distance of approach, kinetic energy is converted into potential energy]

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