Question

In 1911, Ernest Rutherford and his assistants Geiger and Marsden conducted an experiment in which they scattered alpha particles (nuclei of helium atoms) from thin sheets of gold. An alpha particle, having charge +2e and mass 6.64 10-27 kg, is a product of certain radioactive decays. The results of the experiment led Rutherford to the idea that most of an atom's mass is in a very small nucleus, with electrons in orbit around it. Assume an alpha particle, initially very far from a stationary gold nucleus, is fired with a velocity of 1.49 107 m/s directly toward the nucleus (charge +79e). What is the smallest distance between the alpha particle and the nucleus before the alpha particle reverses direction? Assume the gold nucleus remains stationary.

Answer 1

Concepts and reason

The concept required to solve this problem is the conservation of energy.

Substitute the expression of potential energy and kinetic energy in energy conservation principle to finally solve for the expression of distance.

Fundamentals

Kinetic energy: It is defined the as energy due to its motion.

Electric potential energy: It is defined as the energy due to point change of particle.

The expression of kinetic energy can be written as,

$K = \frac{1}{2}m{v^2}$

Here, $K$ is the is the kinetic energy of particle, $m$ is the mass of alpha particle and $v$ is the velocity of alpha particle.

The expression of electric potential energy can be written as,

$U = \frac{{k{q_\alpha }{q_{Au}}}}{r}$

Here, $U$ is the electric potential energy of two particles, $k$ is the electric constant, ${q_\alpha }$ is the charge of alpha particle, ${q_{Au}}$ is the charge of gold and $r$ is the distance between two particles.

The kinetic energy of the $+ 2e$ change (alpha) is fired at $1.49107{\rm{ m/s}}$ toward the nucleus of gold which charge is $+ 79e$ . During this process, the kinetic energy of alpha particle is converted into potential energy. Therefore, energy is nether created nor destroyed during the process and total energy of this system is constant that is conservation of energy.

The conservation of energy of this system can be written as,

${K_i} + {U_i} = {K_f} + {U_f}$

Here, ${K_i}$ and ${K_f}$ are the initial and final kinetic energy of the of alpha particle respectively, ${U_i}$ and ${U_f}$ are the initial and final electric potential energy of the system respectively.

In this above expression, the kinetic energy of the $\alpha$ goes into electric potential energy. Therefore, the initial potential energy of this particle is zero and final kinetic energy of this particle is also zero. Hence the above expression can be written as,

The above expression can be written as,

$\begin{array}{c}\\{K_i} = {U_f}\\\\\frac{1}{2}m{v_i}^2 = \frac{{k{q_\alpha }{q_{Au}}}}{{{r_f}}}\\\end{array}$

Here, subscripts $i$ and $f$ denotes initial and final condition of state.

The distance between alpha particle and nucleus of gold can be calculated as,

$\frac{1}{2}m{v_i}^2 = \frac{{k{q_\alpha }{q_{Au}}}}{{{r_f}}}$

Substitute the value, $+ 2e$ for ${q_\alpha }$ and $+ 79e$ for ${q_{Au}}$ in the above expression,

$\begin{array}{c}\\\frac{1}{2}m{v_i}^2 = \frac{{k\left( { + 2e} \right)\left( { + 79e} \right)}}{{{r_f}}}\\\\{r_f} = \frac{{2k\left( { + 2e} \right)\left( { + 79e} \right)}}{{m{v_i}^2}}\\\\ = \frac{{316k{e^2}}}{{m{v_i}^2}}\\\end{array}$

Here, $e$ is the charge of electron.

Substitute the value ${\rm{9}} \times {\rm{1}}{{\rm{0}}^9}{\rm{ N}}{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}$ for $k$ , $1.60 \times {10^{ - 19}}{\rm{ C}}$ for $e$ , $6.64 \times {10^{ - 27}}{\rm{ kg}}$ for $m$ and $1.49 \times {10^7}{\rm{ m/s}}$ for ${v_i}$ in the above expression.

$\begin{array}{c}\\{r_f} = \frac{{316\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^9}{\rm{ N}}{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right){{\left( {1.60 \times {{10}^{ - 19}}{\rm{ C}}} \right)}^2}}}{{\left( {6.64 \times {{10}^{ - 27}}{\rm{ kg}}} \right){{\left( {1.49 \times {{10}^7}{\rm{ m/s}}} \right)}^2}}}\\\\ = 4.931 \times {10^{ - 14}}{\rm{ m}}\\\\{\rm{ = }}4.931 \times {10^{ - 14}}{\rm{ m}}\left( {\frac{{1{\rm{ fm}}}}{{{{10}^{ - 15}}}}} \right)\\\\ = 49.31{\rm{ fm}}\\\end{array}$

Ans:

The distance between alpha particle and nucleus of gold is $49.31{\rm{ fm}}$ .