Question

In one of the classic nuclear physics experiments at the beginning of the \(20{\rm{ }}th\) century, an alpha particle was accelerated toward a gold nucleus, and its path was substantially deflected by the Coulomb interaction. If the energy of the doubly charged alpha nucleus was \(5.00{\rm{ }}MeV\), how close to the gold nucleus (\(79\) protons) could it come before being deflected?

Short answer

From the gold nucleus the distance of closet approach of the alpha nucleus is obtained as: \(r{\rm{ }} = {\rm{ }}4.55{\rm{ }} \times {\rm{ }}{10^{ - 14}}{\rm{ }}m\).

Step-by-step solution

The given data and the required data

The alpha nucleus is said to be charged double.

The electric potential energy of the alpha nucleus is at a distance of closest approach which is:\(Ue{\rm{ }} = {\rm{ }}5.00{\rm{ }}MeV\).

The gold nucleus has\(79\)protons.

The value of the fundamental charge is:\(e{\rm{ }} = {\rm{ }}1.60{\rm{ }} \times {\rm{ }}{10^{ - 19}}{\rm{ }}C\).

Determining the distance of closet approach of the alpha nucleus from the gold nucleus.

Define Electric Field

The term "electric field" refers to a physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them.

Concepts and Principles

Electron Volt: It is the energy given to a fundamental charge accelerated through a potential difference of\(\left( {1{\rm{ }}V} \right)\). It is in equation form as:

\(\begin{array}{c}1{\rm{ }}eV{\rm{ }} = {\rm{ }}\left( {1.60{\rm{ }}x{\rm{ }}10 - 19{\rm{ }}C} \right)\left( {1{\rm{ }}V} \right)\\ = {\rm{ }}1.60{\rm{ }}x{\rm{ }}10 - 19{\rm{ }}J\end{array}\)…………………….(I)

The electric potential energy associated with a pair of point charges separated by a distance\({\rm{r}}\)is in the form of:

\({U_e}{\rm{ }} = {\rm{ }}k\frac{{{q_1}{q_2}}}{r}\)…………………….(II)

The value of is a proportionality constant known as Coulomb's constant. The unit of electric potential energy is joule \((J)\).

Evaluating the distance of closest approach of alpha from gold nucleus

Converting the electric potential energy of the alpha nucleus\(Ue\)from\(MeV\)to joules using the conversion factor from the first equation as:

\(\begin{array}{c}{U_e}{\rm{ }} = {\rm{ }}(5.00{\rm{ }}MeV)(\frac{{{{10}^6}{\rm{ }}eV}}{{1{\rm{ }}MeV}})(\frac{{1.60{\rm{ }} \times {\rm{ }}{{10}^{ - 19}}{\rm{ }}J}}{{1{\rm{ }}eV}})\\ = {\rm{ }}8.00{\rm{ }} \times {\rm{ }}{10^{ - 13}}{\rm{ }}J\end{array}\)

Electricpotential energy of the system of the alpha nucleus and the gold nucleus at the distance of closest approach is found using the second equation as:

\({U_e}{\rm{ }} = {\rm{ }}k\frac{{{q_\alpha }{\rm{ }}{q_{gold}}}}{r}\)

The value of\({q_\alpha }{\rm{ }} = {\rm{ }}2e\)which is the charge of the alpha nucleus.

The value of\({q_{gold}}{\rm{ }} = {\rm{ }}79e\)which is the charge of the gold nucleus.

The value of\(r\)is the distance of closest approach.

Substituting known values into the above expression as:

\(\begin{array}{c}{U_e}{\rm{ }} = {\rm{ }}k\frac{{(2{\rm{ }}e)(79{\rm{ }}e)}}{r}\\ = {\rm{ }}k\frac{{158{\rm{ }}{e^2}}}{r}\end{array}\)

Solving the value of\(r\)as:

\(r{\rm{ }} = {\rm{ }}k\frac{{158{\rm{ }}{e^2}}}{{{U_e}}}\)

Entering the values and we obtain:

\(\begin{array}{c}r{\rm{ }} = {\rm{ }}8.99{\rm{ }} \times {\rm{ }}{10^9}{\rm{ }}N.{m^2}/C\frac{{158{{(1.60{\rm{ }} \times {\rm{ }}{{10}^{ - 19}}{\rm{ }}C)}^2}}}{{8.00{\rm{ }} \times {\rm{ }}{{10}^{ - 13}}{\rm{ }}J}}\\ = {\rm{ }}4.55{\rm{ }} \times {\rm{ }}{10^{ - 14}}{\rm{ }}m\end{array}\)

Therefore, distance of closet approach of the alpha nucleus from the gold nucleus is: \(r{\rm{ }} = {\rm{ }}4.55{\rm{ }} \times {\rm{ }}{10^{ - 14}}{\rm{ }}m\).