Question

In your laboratory, you set up an experiment with an electron gun that emits electrons with energy of \(7.50 \mathrm{keV}\) toward an atomic target. What deflection (magnitude and direction) would Earth's magnetic field \((0.300 \mathrm{G})\) produce in the beam of electrons if the beam is initially directed due east and covers a distance of \(1.00 \mathrm{~m}\) from the gun to the target? (Hint: First calculate the radius of curvature, and then determine how far away from a straight line the electron beam has deviated after \(1.00 \mathrm{~m}\).)

Short answer

Answer: To find the magnitude of the deflection, follow these steps: 1. Convert the energy to Joules: $E_\text{Joule} = 7.50 \times 10^3 \mathrm{~eV} \times 1.6 \times 10^{-19}\mathrm{~J/eV}$ 2. Find the velocity: $v = \sqrt{\frac{2E_\text{Joule}}{m_\text{electron}}}$ 3. Convert the magnetic field to Tesla: $B_\text{Tesla} = 0.300 \times 10^{-4}\mathrm{~T}$ 4. Find the radius of the circular motion: $r = \frac{m_\text{electron}v}{|q|B_\mathrm{Tesla}}$ 5. Use the law of cosines to find the deflection magnitude: $\text{Deflection}^2 = r^2 + (1.00\mathrm{~m})^2 - 2 r (1.00\mathrm{~m})\cos(\theta)$ The direction of the deflection is south.

Step-by-step solution

Determine the velocity of electrons

First, we will use the given energy of the electrons and convert it into Joules. Then, we will find the velocity using the kinetic energy formula. The given energy of the electrons is 7.50 keV. $$ \begin{aligned} E &= 7.50 \times 10^3 \mathrm{~eV} \end{aligned} $$ To convert the electron-volt to Joule, we will multiply by electron charge in Joule units \((1 \mathrm{~eV} = 1.6 \times 10^{-19}\mathrm{~J})\). $$ \begin{aligned} E_\text{Joule} &= E \times 1.6 \times 10^{-19}\mathrm{~J/eV} \end{aligned} $$ As we have the energy in Joule, the relationship between kinetic energy and velocity is given by $$ \begin{aligned} E_\text{Joule} &= \frac{1}{2}m_\text{electron}v^2 \end{aligned} $$ Where \(m_\text{electron} = 9.11 \times 10^{-31}\mathrm{~kg}\) is the mass of the electron. Solving for the velocity, we get $$ \begin{aligned} v &= \sqrt{\frac{2E_\text{Joule}}{m_\text{electron}}} \end{aligned} $$

Calculate the magnetic force acting on the electrons

The magnetic force acting on the electrons is given by $$ \begin{aligned} F_\text{mag} &= qvB \end{aligned} $$ Where \(q\) is the electron charge, \(v\) is the electron velocity, and \(B\) is the magnetic field strength. The strength of Earth's magnetic field is given as \(0.300\mathrm{~G}\), which we will need to convert to Tesla (1 Gauss = \(10^{-4}\) Tesla). $$ \begin{aligned} B_\text{Tesla} &= 0.300 \times 10^{-4}\mathrm{~T} \end{aligned} $$

Determine the radius of the circular motion due to the magnetic force

The magnetic force acting on the electrons will cause them to move in a circular path. The centripetal force for the circular motion will be equal to the magnetic force. $$ \begin{aligned} F_\text{centripetal} = F_\text{mag} &= \frac{m_\text{electron}v^2}{r} \end{aligned} $$ Where \(r\) is the radius of the circular path. To find the radius, we can rearrange the equation as follows: $$ \begin{aligned} r &= \frac{m_\text{electron}v}{|q|B_\mathrm{Tesla}} \end{aligned} $$

Determine the deflection (magnitude and direction) after the electrons travel 1.00m

To find the deflection (magnitude), we can use the equation for a right triangle, with its hypotenuse as the radius \(r\), and one side along the initially straight path as \(1.00\mathrm{~m}\). The angles between \(1.00\mathrm{~m}\) and the hypotenuse (at the target and electron gun) are the same, so let this angle be \(\theta\). Using the law of cosines, we can find the length of the other side, which is the deflection: $$ \begin{aligned} \text{Deflection}^2 &= r^2 + (1.00\mathrm{~m})^2 - 2 r (1.00\mathrm{~m})\cos(\theta) \end{aligned} $$ As the angle between the magnetic field and the velocity is \(90^\circ\), the magnetic field is pointing north. Since electrons are negatively charged, they will be deflected to the south. Thus, the direction of the deflection is south.

Key concepts

Electron Gun Experiment

In the electron gun experiment, electrons are emitted towards a target as beams from an electron gun. These electrons have a certain amount of energy given in electron-volts (eV). In our example, the energy is 7.50 keV.

When we deal with problems like these, one of the first steps is converting the energy from electron-volts to Joules. This is essential because most physics formulas, such as those for kinetic energy, use Joules.
This conversion is performed by multiplying the energy in eV by the electron charge, which is approximately 1.6 × 10^{-19} J/eV.

• An electron gun accelerates electrons to a specific energy level.
• Converting units is crucial to apply them in equations.

Understanding how to handle and manipulate these small units of energy is vital for working with and interpreting experiments involving particles like electrons.

Earth's Magnetic Field Impact

The Earth's magnetic field can influence the path of charged particles, including electrons. This magnetic field is measured in Gauss (G) but needs to be converted to Tesla (T) to be used in calculations related to magnetic force.

Earth's field is roughly 0.300 G which would be equal to 0.300 x 10^{-4} T. In an electron gun experiment, the electrons move at a velocity determined by their kinetic energy, encountering this magnetic field along their path.

• Magnetic force on a moving charge is calculated by the equation: \( F_{mag} = qvB \)
• q is the electron charge, v is the velocity, and B is the magnetic field in Tesla.

The magnetic force will cause these electrons to deflect from their path due to the Lorentz force, making it imperative to account for the Earth's magnetic field in experiments.

Curved Electron Path

When electrons travel under the influence of a magnetic field, they experience a force perpendicular to their velocity and the direction of the magnetic field — this is a result of the Lorentz force. Because of this perpendicular force, electrons travel along a curved path rather than a straight line.

The curvature of the path causes the electrons to move in a circular trajectory. The radius of this circle is determined by the balance of the magnetic force and the centripetal force required for circular motion.

• The formula \( F_{centripetal} = \frac{m_{electron}v^2}{r} \) helps find the radius.
• It equals the magnetic force \( qvB \), which allows calculation of r.

Understanding this curvature is key to determining the path of electrons and predicting where they will strike the target in experiments.

Kinetic Energy Conversion

In physics, kinetic energy conversion is a common calculation required to find the velocity of particles, such as electrons from an electron gun. Starting with calculated energy in Joules, kinetic energy formula comes into play: \( E_{joule} = \frac{1}{2}m_{electron}v^2 \) To find velocity, you rearrange this to solve for v, providing insights into how fast the electrons are moving.

Understanding the electron's speed is vital for further computations, like determining the effect of external fields, such as Earth's magnetic field.

• Kinetic energy aids in calculating initial velocity.
• Converting energy correctly ensures the right values in equations.

This process links energy and motion, serving as a basis for exploring deeper physical phenomena.