Problem 61 - Fundamentals of Physics

Question

A steady beam of alpha particles \((q=+2 e)\) traveling with constant kinetic energy \(20 \mathrm{MeV}\) carries a current of \(0.25 \mu \mathrm{A}\). (a) If the beam is directed perpendicular to a flat surface, how many alpha particles strike the surface in \(3.0 \mathrm{~s} ?\) (b) At any instant, how many alpha particles are there in a given \(20 \mathrm{~cm}\) length of the beam? (c) Through what potential difference is it necessary to accelerate each alpha particle from rest to bring it to an energy of \(20 \mathrm{MeV} ?\)

Step-by-Step Solution

Answer

(a) Approximately \(2.34 \times 10^{12}\) particles. (b) About \(4.88 \times 10^5\) particles. (c) 10 million volts.

Step 1: Calculate charge of one alpha particle

Since an alpha particle carries a charge of +2e and the elementary charge e is approximately \(1.6 \times 10^{-19}\) C, we calculate the charge of one alpha particle as:\[q = 2 \times 1.6 \times 10^{-19}\, \text{C} = 3.2 \times 10^{-19}\, \text{C}\]

Step 2: Find number of particles per time period

The relationship between current (I), charge (q), and number of particles (N) can be expressed as \(N = \frac{I \cdot t}{q}\). Here, \(I = 0.25 \times 10^{-6}\) A, \(t = 3\) s, and \(q\) is the charge of one alpha particle calculated in Step 1.\[N = \frac{0.25 \times 10^{-6} \times 3}{3.2 \times 10^{-19}} = 2.34 \times 10^{12}\]Thus, approximately \(2.34 \times 10^{12}\) alpha particles strike the surface in \(3\) seconds.

Step 3: Calculate speed of alpha particles

The kinetic energy (KE) is given by the formula \(\text{KE} = \frac{1}{2} mv^2\), where \(m\) is the mass of the alpha particle (approximately \(6.64 \times 10^{-27}\) kg). Solve for \(v\) given \(\text{KE} = 20\) MeV = \(20 \times 10^6 \times 1.6 \times 10^{-13}\) J:\[20 \times 10^6 \times 1.6 \times 10^{-13} = \frac{1}{2} \times 6.64 \times 10^{-27} \times v^2 \]\[v = \sqrt{\frac{2 \times 20 \times 10^6 \times 1.6 \times 10^{-13}}{6.64 \times 10^{-27}}} \approx 1.55 \times 10^7 \text{ m/s}\]

Step 4: Calculate the number of particles in 20 cm segment

The number of particles in a length \(L\) of the beam can be found from the speed \(v\) calculated in Step 3 using the formula \(\frac{I}{q} = \frac{N \cdot v}{L}\), solve for \(N\) when \(L = 0.2\) m:\[N = \frac{I \cdot L}{q \cdot v} = \frac{0.25 \times 10^{-6} \times 0.2}{3.2 \times 10^{-19} \times 1.55 \times 10^7} \approx 4.88 \times 10^5\]Therefore, there are about \(4.88 \times 10^5\) alpha particles in any given \(20\) cm segment.

Step 5: Determine the potential difference required

The energy gained by an alpha particle when accelerated through a potential difference \(V\) is \(qV = \text{KE}\). Thus:\[V = \frac{\text{KE}}{q} = \frac{20 \times 10^6 \times 1.6 \times 10^{-13}}{3.2 \times 10^{-19}} = 10 \times 10^6 \text{ V} = 10 \text{ MV}\]Hence, a potential difference of \(10\) million volts is needed.

Key Concepts

Charge of Alpha Particle

An alpha particle is a helium nucleus composed of two protons and two neutrons. It carries a positive charge due to its protons. The charge of an alpha particle is twice the elementary charge of an electron, denoted as "+2e". The elementary charge, denoted by "e", is a fundamental physical constant in nature, approximately equal to \(1.6 \times 10^{-19}\) coulombs. To find the charge of a single alpha particle, we simply multiply the elementary charge by two:

• Charge of alpha particle: \(q = 2 \times 1.6 \times 10^{-19}\, \text{C} = 3.2 \times 10^{-19}\, \text{C}\).

This charge is crucial for calculations involving electric fields and forces acting on the particles, as well as any potential difference or current calculations.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. For alpha particles in motion, this energy plays a key role in various calculations. The kinetic energy is given by the formula:

• \(\text{KE} = \frac{1}{2} mv^2\)

where "m" stands for the mass of the alpha particle and "v" is its velocity.In many physics problems, kinetic energy will be given in electronvolts (eV) or mega-electronvolts (MeV). Here, the kinetic energy was stated to be \(20\, \text{MeV}\). To convert it to joules (J), use the conversion factor where \(1\, \text{eV} = 1.6 \times 10^{-19}\, \text{J}\):

• \(20 \times 10^6 \times 1.6 \times 10^{-13}\, \text{J}\)

Understanding kinetic energy helps in determining how fast an alpha particle is moving and how these particles interact with electric fields.

Potential Difference

The potential difference is an important concept when discussing the behavior of charged particles. When a charged particle moves through a potential difference, it gains or loses energy. For alpha particles, the potential difference is necessary to accelerate them from rest to a specified kinetic energy level.The relationship between potential difference (\(V\)), charge (\(q\)), and kinetic energy (\(\text{KE}\)) is:

• \(qV = \text{KE}\)

To find the potential difference required to accelerate the alpha particles:

• \(V = \frac{\text{KE}}{q}\)

Given the initial kinetic energy of \(20\, \text{MeV}\) and the charge \(q = 3.2 \times 10^{-19}\, \text{C}\), the potential difference is:

• \(V = 10\, \text{MV}\)

This means a potential difference of 10 million volts is required to reach the desired kinetic energy.

Current Calculation

Current is the flow of electric charge and in the context of an alpha particle beam, it tells us the rate at which these particles pass through a surface.The formula relating current (\(I\)), charge (\(q\)), and time (\(t\)) to the number of particles (\(N\)) is:

• \(N = \frac{I \cdot t}{q}\)

For the given scenario:

• \(I = 0.25 \times 10^{-6}\, \text{A}\)
• \(t = 3\, \text{s}\)
• \(q = 3.2 \times 10^{-19}\, \text{C}\)

Substituting the values, the number of alpha particles striking the surface is approximately \(2.34 \times 10^{12}\). This calculation is fundamental for understanding how many particles interact with a specific area over time, providing insight into both theoretical and practical aspects of the beam's impact.