49 Ernest Rutherford the first New Zealander to be awarded t

49. 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 (4 He) from gold-197 nuclei (197Au). The energy of the incoming helium nucleus was 8.00×1013J, and the masses of the helium and gold nuclei were 6.68×1027kg and 3.29×1025kg, respectively (note that their mass ratio is 4 to 197). (a) If a helium nucleus scatters to an angle of 120º during an elastic collision with a gold nucleus, calculate the helium nucleus’s final speed and the final velocity (magnitude and direction) of the gold nucleus. (b) What is the final kinetic energy of the helium nucleus?

Solution

Inital Kinetic Energy = 8.00 * 10^-13 J
Let the initial Speed of Helium nucleus = v1

0.5 * mh * v1^2 = 8.00 * 10^-13 J
v1 = sqrt[(2 * 8.00 * 10^-13)/(6.68 * 10^-27)]
v1 = 1.548 * 10^7 m/s

Conservation of momentum along the x-axis gives:
mh * v1 = mh *v2*sin(30) + mg*vg*cos()
6.68 * 10^-27 * 1.548 * 10^7 = 6.68 * 10^-27 * v2 * sin(30) +  3.29 *10^25 * vg*cos() -------1

Conservation of momentum along the y-axis gives:
0 =  mh *v2*cos(30) - mg*vg*sin()
0 =  6.68 * 10^-27 * v2 * cos(30) +  3.29 *10^25 * vg*sin() -----------2

Using Energy Conservation -
Inital Kinetic Energy = Final Kinetic Energy
8.00 * 10^-13 J =  0.5*mh *vh^2 + 0.5*mg *vg^2
16.0 * 10^-13 =  6.68 * 10^-27 * vh^2 + 3.29 *10^25 * vg^2 ----------3

Using eq 1 & 2 & 3
Solving for vh , vg and
vh = 1.5 * 10^7 m/s
vg = 5.36 * 10^5 m/s
= - 29.5^o (-ve sign means below the Horizontal)

(A)
Final Velocity, vh = 1.5 * 10^7 m/s
Direction, = - 29.5^o (-ve sign means below the Horizontal)

(B)
Final Kinetic Energy of the Helium Nucleus -
K.Efin = 0.5 * mh * vh^2
K.Efin = 0.5 * 6.68 * 10^-27  *  (1.5 * 10^7)^2
K.Efin = 7.52 * 10^-13 J

Final Kinetic Energy of the Helium Nucleus, K.Efin = 7.52 * 10^-13 J