Question

Ernest Rutherford (the first New Zealander to be awarded the Nobel rize in Chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei\(\left( {{}^4He} \right)\)from gold-197 nuclei\(\left( {{}^{197}Au} \right)\).The energy of the incoming helium nucleus was\(8.00 \times {10^{ - 13}}\;{\rm{J}}\), and themasses of the helium and gold nuclei were\(6.68 \times {10^{ - 27}}\;{\rm{kg}}\)and\(3.29 \times {10^{ - 25}}\;{\rm{kg}}\), respectively (note that their mass ratio is 4 to 197). (a) If a helium nucleus scatters to an angle of\(120^\circ \)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?

Short answer

1. The final velocity is\(5.36 \times {10^5}\;{\rm{m/s}}\)at an angle\({\theta _2} = - 29.5^\circ \).
2. Kinetic energy of He nucleus is\(7.53 \times {10^{ - 13}}\;{\rm{J}}\)

Step-by-step solution

Definition of Final Velocity

A final velocity is defined as the final speed of a moving object with an initial velocity and acceleration over some time.

The initial velocity of Helium is,

\(\begin{aligned}{u_{He}} = \sqrt {\dfrac{{2{K_{He}}}}{{{m_{He}}}}} \\ = \sqrt {\dfrac{{2 \times 8 \times {{10}^{ - 13}}}}{{6.68 \times {{10}^{ - 27}}}}} \\ = 1.54 \times {10^7}\;{\rm{m/s}}\end{aligned}\)

From the conservation of momentum along both the axes we get,

\(\begin{aligned}{{m}_{{He}}}{{u}_{{He}}}{ = }{{m}_{{He}}}{{v}_{{He}}}{cos}{{\theta }_{{He}}}{ + }{{m}_{{Au}}}{{v}_{{Au}}}{cos}{{\theta }_{{Au}}}\;....{(1)}\\{0 = }{{m}_{{He}}}{{v}_{{He}}}{sin}{{\theta }_{{He}}}{ + }{{m}_{{Au}}}{{v}_{{Au}}}{sin}{{\theta }_{{Au}}}\;....{(2)}\end{aligned}\)

Applying conservation of kinetic energy we get,

\(\begin{aligned}\dfrac{1}{2}{m_{He}}{u_{He}}^2 = \dfrac{1}{2}{m_{He}}{v_{He}}^2 + \dfrac{1}{2}{m_{Au}}{v_{Au}}^2\\{v_{Au}}^2 = \dfrac{{{m_{He}}\left( {{u_{He}}^2 - {v_{He}}^2} \right)}}{{{m_{Au}}}}\end{aligned}\)

Combining these equations we get,

\(\left( {{{m}_{{He}}}^{2}{ + }{{m}_{{Au}}}{{m}_{{He}}}} \right){{v}_{{He}}}^{2}{ - }\left[ {{2}{{m}_{{He}}}^{2}{{u}_{{He}}}{cos}{{\theta }_{{He}}}} \right]{{v}_{{He}}}{ = }\left( {{{m}_{{Au}}}{{m}_{{He}}}{ - }{{m}_{{He}}}^{2}} \right){{u}_{{He}}}^{2}\)

Substituting the values,

\(\begin{aligned}\left[ {{{\left( {{6}{.68 \times 1}{{0}^{{ - 27}}}} \right)}^{2}}{ + }\left( {{3}{.29 \times 1}{{0}^{{ - 23}}}} \right)\left( {{6}{.68 \times 1}{{0}^{{ - 27}}}} \right)} \right]{{v}_{{He}}}^{2}{ - }\left[ {{2 \times }{{\left( {{6}{.68 \times 1}{{0}^{{ - 27}}}} \right)}^{2}}\left( {{1}{.54 \times 1}{{0}^{7}}} \right){cos120^\circ }} \right]{{v}_{{He}}}\\{ = }\left[ {\left( {{3}{.29 \times 1}{{0}^{{ - 23}}}} \right)\left( {{6}{.68 \times 1}{{0}^{{ - 27}}}} \right){ - }{{\left( {{6}{.68 \times 1}{{0}^{{ - 27}}}} \right)}^{2}}} \right]{\left( {{1}{.54 \times 1}{{0}^{7}}} \right)^{2}}\end{aligned}\)

The velocity is,

\(\begin{aligned}{{v}_{{He}}}{ = }\dfrac{{{ - }\left( {{6}{.90599 \times 1}{{0}^{{ - 46}}}} \right){ + }\sqrt {{{\left( {{6}{.90599 \times 1}{{0}^{{ - 46}}}} \right)}^{2}}{ + 4}\left( {{2}{.24234 \times 1}{{0}^{{ - 51}}}} \right)\left( {{5}{.1571 \times 1}{{0}^{{ - 37}}}} \right)} }}{{{2}\left( {{2}{.34234 \times 1}{{0}^{{ - 51}}}} \right)}}\;{m/s}\\{ = 1}{.50 \times 1}{{0}^{7}}\;{m/s}\end{aligned}\)

Determine the angle

The final velocity of gold is,

\(\begin{array}{c}{v_{Au}} = \sqrt {\frac{{\left( {6.68 \times {{10}^{ - 27}}} \right){{\left( {1.547 \times {{10}^7}} \right)}^2} - {{\left( {1.50 \times {{10}^7}} \right)}^2}}}{{3.29 \times {{10}^{ - 25}}}}} \;{m/s}\\ = 5.35 \times {10^5}\;{\rm{m/s}}\end{array}\)

The angle of the gold is,

\(\begin{array}{c}\sin {\theta _{Au}} = \frac{{{m_{He}}{v_{He}}\sin {\theta _{He}}}}{{{m_{Au}}{v_{Au}}}}\\\sin {\theta _{Au}} = \frac{{\left( {6.68 \times {{10}^{ - 27}}} \right)\left( {1.547 \times {{10}^7}} \right)\sin 120^\circ }}{{\left( {3.29 \times {{10}^{ - 25}}} \right)\left( {5.3 \times {{10}^5}} \right)}}\\{\theta _{Au}} = 29.7^\circ \end{array}\)

Determine the kinetic energy

The kinetic energy of He is,

\(\begin{aligned}{}\frac{1}{2}{m_{He}}{v_{He}}^2\\ &= \frac{1}{2} \times \left( {6.68 \times {{10}^{ - 27}}} \right){\left( {1.547 \times {{10}^7}} \right)^2}\;{\rm{J}}\\ &= 7.53 \times {10^{ - 13}}\;{\rm{J}}\end{aligned}\)