Question

A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in "head-on" to a particular lead nucleus and stops 6.50 $\times$ 10${}^{-14}$ m away from the center of the nucleus. (This point is well outside the nucleus.) Assume that the lead nucleus, which has 82 protons, remains at rest. The mass of the alpha particle is 6.64 $\times$ 10${}^{-27}$ kg. (a) Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in joules and in MeV. (b) What initial kinetic energy (in joules and in MeV) did the alpha particle have? (c) What was the initial speed of the alpha particle?

Short answer

(a) 3.17 × 10^{-12} J, 19.8 MeV; (b) same as (a); (c) 1.23 × 10^7 m/s.

Step-by-step solution

Identify Components and Parameters

The problem involves the interaction of an alpha particle with a lead nucleus. We know:- The distance to stop, $r=6.50\times {10}^{-14}$ m.- The charge of the lead nucleus is $Z=82$.- The charge of an alpha particle $=2e$, where $e=1.60\times {10}^{-19}$ C.- Mass of alpha particle $=6.64\times {10}^{-27}$ kg.

Calculate Electrostatic Potential Energy in Joules

The electrostatic potential energy $U$ between two charges can be calculated using the formula:$$U=\frac{k\cdot q_1\cdot q_2}{r}$$where:- $k=8.99\times {10}^9{\text{Nm}}^2/{\text{C}}^2$ is the Coulomb's constant.- $q_1=2e$ the charge of alpha particle, $q_2=82e$ the charge of lead nucleus.Substituting, we get:$$U=\frac{(8.99\times {10}^9)\cdot (2\times 1.60\times {10}^{-19})\cdot (82\times 1.60\times {10}^{-19})}{6.50\times {10}^{-14}}\approx 3.17\times {10}^{-12}\text{J}$$

Convert Joules to MeV

The conversion factor between joules and MeV is $1\text{J}=6.242\times {10}^{12}\text{MeV}$. Thus,$$U_{\text{MeV}}=3.17\times {10}^{-12}\text{J}\times 6.242\times {10}^{12}\text{MeV/J}\approx 19.8\text{MeV}$$

Determine Initial Kinetic Energy

The alpha particle's initial kinetic energy is equal to its potential energy at the stopping point due to conservation of energy:$$K{E}_{\text{initial}}=U=3.17\times {10}^{-12}\text{J}\approx 19.8\text{MeV}$$

Calculate Initial Speed

Use the kinetic energy formula:$$KE=\frac{1}{2}m{v}^2$$Rearranging gives$$v=\sqrt{\frac{2\cdot KE}{m}}$$Substitute $KE=3.17\times {10}^{-12}\text{J}$ and $m=6.64\times {10}^{-27}\text{kg}$:$$v=\sqrt{\frac{2\cdot 3.17\times {10}^{-12}}{6.64\times {10}^{-27}}}\approx 1.23\times {10}^7\text{m/s}$$

Key concepts

Alpha Particle

An alpha particle is essentially a helium nucleus. It consists of two protons and two neutrons, making it a positively charged particle.

• The charge of an alpha particle is due to its two protons, each carrying a charge of $+e$, where $e=1.60\times {10}^{-19}\text{ C}$.
• In this scenario, the total charge is $2e$ because there are two protons.
• Its mass is significantly larger than that of a single proton, calculated at $6.64\times {10}^{-27}\text{ kg}$.

Alpha particles are often involved in nuclear reactions and are commonly encountered in alpha decay. Due to their positive charge, they are affected by electric fields and, as in this scenario, will interact with other charged particles through electrostatic forces.

Coulomb's Law

Coulomb's Law describes the electrostatic interaction between charged particles. It states that the force between two charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them.For potential energy, the formula is similar:$$U=\frac{k\cdot q_1\cdot q_2}{r}$$

• $k,$ Coulomb's constant, is approximately $8.99\times {10}^9{\text{Nm}}^2/{\text{C}}^2$.
• $q_1$ and $q_2$ represent the charges of the interacting particles.
• $r$ is the distance between the centers of the two charges.

In our case, you have the alpha particle and a lead nucleus. When these two get close, they exert significant forces on each other, and as they do so, they also have potential energy between them, calculated using this formula.

Initial Kinetic Energy

The initial kinetic energy of the alpha particle needs to be such that when entirely converted to potential energy, it brings the particle to a stop at a particular distance.

• The concept of kinetic energy here is essential for understanding the particle's motion and how much energy is needed to reach that 'stop point.'
• This initial kinetic energy value is equivalent to the potential energy at the closest approach due to the conservation of energy.

Kinetic energy gives us a picture of the particle's initial state in terms of motion. When the problem mentions 'initial kinetic energy,' it's highlighting how much energy the particle begins with as it travels towards the lead nucleus.

Conversion of Energy

Conversion of energy principles state that energy cannot be created or destroyed, only transformed from one form to another. In the problem context, you have kinetic energy being transformed into potential energy.

• As the alpha particle approaches the lead nucleus, its kinetic energy decreases.
• This decrease corresponds to an increase in electrostatic potential energy between the alpha and the lead nucleus.
• When all kinetic energy is converted, that's where the particle comes to rest or stops, showing maximum potential energy.

Conservation principles help in understanding various exercises in physics, particularly when dealing with isolated systems, like the alpha particle and nucleus system here.

Kinetic Energy Formula

The kinetic energy formula is a fundamental concept in physics. It is given by:$$KE=\frac{1}{2}m{v}^2$$

• $KE$ represents the kinetic energy of a particle or object.
• $m$ is the mass of the object, and $v$ is its velocity.

This formula relates directly to our problem since it lets us determine the initial speed of the alpha particle. By rearranging the formula, you can solve for the speed $v$:$$v=\sqrt{\frac{2\cdot KE}{m}}$$This rearrangement comes into play after determining the initial kinetic energy, as it allows you to calculate how fast the alpha particle was moving initially. It connects the particle's motion with its energy state.