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 (4He) from gold-197 nuclei (197Au). The energy of the incoming helium nucleus was 7.34 ✕ 10−13 J, and the masses of the helium and gold nuclei were 6.68 ✕ 10−27 kg and 3.29 ✕ 10−25 kg, respectively (note that their mass ratio is 4 to 197). (Assume that the helium nucleus travels in the +x direction before the collision.) (a) If a helium nucleus scatters to an angle of 132° 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 counterclockwise from the +x-axis) of the gold nucleus. (b) What is the final kinetic energy (in J) of the helium nucleus?

Answer 1

(a) The formula for kinetic Energy (K.E) is:

where;

= mass of the helium

= velocity of helium