Question

IP An α particle with a kinetic energy of 0.45 MeV approaches a stationary gold nucleus.

A)What is the speed of the α particle? (To obtain the mass of an alpha particle, subtract the mass of two electrons from the mass of 4/2He. ) Express your answer using two significant figures.

v= ___ m/s

B)What is the distance of closest approach between the αα particle and the gold nucleus?

d= ___pm

C)If this same αα particle were fired at a copper nucleus instead, would its distance of closest approach be greater than, less than, or the same as that found in part B?

a)The distance of closest approach would be greater than that found in part B.

b)The distance of closest approach would be less than that found in part B.

c)The distance of closest approach would be the same as in part B.

Answer 1

Solution:

A) Kinetic energy of the alpha partice is

An alpha particle consists of 2 protons and 2 neutrons. Mass of an alpha particle is equal to the mass of 2 protons plus mass of 2 neutrons, which is

  

Let v be the speed of alpha particle.

Therefore,

So,

B)

Charge on alpha particle

Charge on gold nucleus

The distance of closest approach is given by,

  

  

(C) (b)

For copper atomic number is 29. The nuclear charge will be

   which is less than Gold nucleus charge.

Since , The distance of closest approach would be less than that of Gold in part B.