Question

An alpha nucleus of energy 1/2 mv² bombards a heavy
nuclear target of charge Ze. Then the distance of closest
approach for the alpha nucleus will be proportional to
(a)1/Ze (b) v² (c) 1/m (d) 1/v⁴

Answer 1

Answer:

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Explanation:

Let r be the distance of closest approach.

At the closest approach, all the kinetic energy of alpha nucleus has converted into the potential energy i.e. K.E=P.E

∴

2

1

mv

2

=

r

kZe

2

We get, r=

mv

2

2kZe

2

⟹r∝

m

1

Answer 2

Explanation:

applies only two electron

(c) the frame in which the electron is at rest is not inert) the motion cl good of