Question

The following charged particles accelerated from rest, through the same potential difference, are projected towards gold nucleus in different experiments. The distance of closest approach will be maximum for

• A. $\alpha -$particles
• B. proton
• C. deuteron
• D. neutron

Answer 1

The correct option is D neutron

Distance of closest approach r $=\frac{q.Q}{4\pi {\epsilon }_0\times K.E}$

Here Q is the charge of the gold foil

The only q is different, rest everything is constant

So, we can say that

r is directly proportional to q

The value of q for :

proton $=1.6\times {10}^{-19}$

neutron = 0

deuteron $=1.6\times {10}^{-19}$

alpha particle $=2\times 1.6\times {10}^{-19}=3.2\times {10}^{-19}$

Since the value of q is greater for alpha particle so the value of r is greater for alpha particle and it will be minimum for the neutron.