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 7.97 ✕ 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 116° 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)

Kinetic energy, E = 0.5 m1 v1^2

Velocity, v1 = sqrt(7.97 x 10^-13/6.68 x 10^-27) = 1.548 x 10^7 m/s

Using law of conservation of momentum,

m1v1 - m1 v1' sin(theta_1) = m2 v2' sin(theta_2)

-m1 v1' sin(theta_1) = m2 v2' sin(theta_2)

Squaring both sides and add the expression,

m2^2 v2'^2 = m1^2 v1 - 2 m1^2 v1 v1'^2 cos(theta_1)+ m1^2 v1'

V2'^2 = (m1^2/m2^2)[v1^2 - 2 v1 v1' cos(theta_1) + v1'^2]

[1 + (6.68 x 10^-27/3.29 x 10^-25)]v1'^2 - [2(6.68 x 10^-27/3.29 x 10^-25)(1.548 x 10^7 cos(116)] v1' - [1 - (6.68 x 10^-27/3.29 x 10^-25)(1.548 x 19^7)^2] = 0

1.02 v^2 - (3.14 x 10^5)v - 2.348 x 10^14 = 0

v1' = 1.50 x 10^7 m/s

(6.68 x 10^-27/3.29 x 10^-25)[(1.548 x 10^7)^2 - (1.5 x 10^7)^2] = v2'^2

V2' = 5.34 x 10^5 m/s

Sin(theta_2) = (6.68 x 10^-27 x 1.5 x 10^7 sin(116))/(3.29 x 10^-25 x 5.34 x 10^5) = - 0.4946

Theta_2 = 29.32°

B)

Energy, E = 0.5 x 6.68 x 10^-27 x (1.5 x 10^7)^2

E = 7.52 x 10^-13 J

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