Question

Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei from gold-197 nuclei. The energy of the incoming helium nucleus was 8.00\times 10^{-13}~\text{J}8.00×10 ​−13 ​​ J, and the masses of the helium and gold nuclei were 6.68\times 10^{-27}~\text{kg}6.68×10 ​−27 ​​ kg and 3.29\times 10^{-25}~\text{kg}3.29×10 ​−25 ​​ kg, respectively (note that their mass ratio is 4 to 197). If a helium nucleus scatters to an angle of 120^\circ120 ​∘ ​​ during an elastic collision with a gold nucleus, what is the final kinetic energy of the helium nucleus?

Answer 1

Answer:

The final kinetic energy of the Helium nucleus (alpha particle) after been scattered through an angle of 120° is

8.00 x 10-13J

Explanation:

In Rutherford Scattering experiment, the collision of the helium nucleus with the gold nucleus is an ELASTIC COLLISION. This means that the kinetic energy is conserved ( The same before and after the collision).

Thus, the final kinetic energy of the helium nucleus is the same as initial kinetic energy (8.00 x 10^-13Joules)

Although, the kinetic energy is converted to potential energy in Coulomb's law equation.

That is,

1/2(mv^2) = (K* q1q2)/r

Where m is the mass of helium nucleus, v is its colliding velocity, k is electrostatic constant, q1 is the charge on helium nucleus, q2 is the charge on gold nucleus, r is impact parameter