Home
Class 12
PHYSICS
A protom of energy 1 MeV is incident hea...

A protom of energy 1 MeV is incident head-on on a gold nucleus (Z = 79), and is scattered through an angle of `180^(@)`. Calculate the distance of nearest approach.

Promotional Banner

Similar Questions

Explore conceptually related problems

An alpha -particle having kinetic energy of 7.5 MeV is scattered by gold (Z=79) mucleus through 180^(@) . Calculate the distance of closest approach.

An alpha -particle of kinetic energy 7.68 MeV is projected towards the nucleus of copper (Z=29). Calculate its distance of nearest approach.

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

A 1 MeV proton is sent against a gold leaf (Z=79) . Calculate the distance of closest approach for head-on collision.

An alpha -particle having K.E. =7.7MeV is scattered by gold (z=79) nucleus through 180^(@) Find distance of closet approach.

An alpha -particle having K.E. =7.7MeV is scattered by gold (z=79) nucleus through 180^(@) Find distance of closet approach.