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Expert-verified Found in: Page 291 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # 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 the masses 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?

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$$7.53 \times {10^{ - 13}}\;{\rm{J}}$$
See the step by step solution

## Step 1: 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}

## Step 2: 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,

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

## Step 3: 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} ### Want to see more solutions like these? 