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?

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

learn it on the intertant so you can solve

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle