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 (He) from gold-197 nuclei (197Au). The energy ofthe incoming helium nucleus was 7.34 x 10-13 J, and the masses of the helium and gold nuclei were 6.68 x 10-27 kg and 3.29 × 10-25 kg, respectively (note that their mass ratio is 4 to 197. Assume that the heliumnucleus travels in the +x-direction before the collision.)(a) If a helium nucleus scatters to an angle of 145° during an elastic collision with a gold nucleus, calculate the helium nucleus' final speed (in m/s) and the final velocity (magnitude in m/s and direction counterclockwisefrom the +x-axis) of the gold nucleus.He nucleus4He speed197Au velocity197 Au directionm/sm/s° counterclockwise from the +x-axis(b) What is the final kinetic energy (in J) of the helium nucleus?120°Gold nucleus

Given that The energy of incoming helium nucleus are E = 7.34×10-13JBut we know E = 12mHevHe2where…

Answered: Ernest Rutherford (the first New…

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