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 7.59 ✕ 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 141° 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. 4He speed Incorrect: Your answer is incorrect. Your answer cannot be understood or graded. More Information m/s 197Au velocity m/s 197Au direction ° counterclockwise from the +x-axis (b) What is the final kinetic energy (in J) of the helium nucleus? Joules

Answer 1

using equilibrium

m₁ V₁ = m₁ V₁ ' + m₂ V₂

the velocity of the Helium

V = ( 2U/m )^1/2

= ( 2 X 7.59 X 10^-13 / 6.68 X 10^-27 )^1/2

V = 15.074 X 10⁶ m.s

6.68 X 10^-27 X 15.074 X 10⁶ i = 6.68 X 10^-27 X 12900000 ( - cos60)i - sin60j ) + 3.29 X 10⁻²⁵ V₂

1 X 10^-19 i = 4.3 X 10^-20 i + 7.46 X 10^-20 j + 3.29 X 10⁻²⁵ V₂

5.7 X 10^-20 i + 7.46 X 10^-20 j = 3.29 X 10⁻²⁵ V₂   

V₂ = 173252.27 i + 226747.72 j

V₂ = ( 173252.27² + 226747.72² )^1/2

V₂ = 285360.96 m/s

tan = 226747.72/ 173252.27

= 52.16^o counter clockwise.