A) Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in c

Question

A) 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 (197Au). See the figure below. The energy of the incoming helium nucleus was 7.70 10-13 J, and the masses of the helium and gold nuclei were 6.68 10-27 and 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° during an elastic collision with a gold nucleus, calculate the helium nucleus's final speed.

? m/s

B) Calculate the final velocity (magnitude and direction) of the gold nucleus. (Assume the positive x direction is the direction in which the helium nucleus is initially traveling, and that it scatters 120° clockwise from the +x-axis.)

Explanation / Answer

initial speed of the He is u1

0,5*m1*u1^2 = 7.7*10^-13 J

u1 = 1.51*10^7 m/s

Applylawof conservation of momentum

along X-axis

m1*u1 = m1*v1*cos(theta) + m2*v2*cos(theta)

6.68*10^-27*1.51*10^7 = -6.68*10^-27*V1*cos(60) + 3.29*10^-25*v2*cos(phi)

10.08*10^-20= -3.34*10^-27*V1 + 3.29*10^-25*v2*cos(phi)...............(1)

along Y-axis

m1*v1*sin(60) = m2*v2*sin(phi)

6.68*10^-27*v1*0.866 = 3.29*10^-25*v2*sin(phi)

v1 = 56.87*v2*sin(phi)...(2)

Substituting (2) in (1) we get

10.08*10^-20= -3.34*10^-27*56.87*v2*sin(phi) + 3.29*10^-25*v2*cos(phi)

10.08*10^-20 = -1.89*10^-25*v2*sin(phi) + 3.29*10^-25*v2*cos(phi)

Apply law of conservation of energy

7.7*10^-13 = 0.5*6.68*10^-27*v1^2 + 0.5*3.29*10^-25*v2^2

7.7*10^-13 = 3.34*10^-27*v1^2 + 1.645*10^-25*v2^2

0.877*10^-6 = 0.0577*10^-12*v1 + 0.405*10^-12*v2

0.877*10^6 = 0.0577*v1 + 0.405*v2

v1 = 15.2*10^6 - 7v2

v1 = 56.87*v2*sin(phi)

then

15.2*10^6 - 7v2 = 56.87*v2*sin(phi)

v2 = 15.2*10^6/(56.87*sin(phi) + 7)

Solve all these equations to get v1 ,v2 and phi

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