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) will 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

Solution: Step 1: Calculate the electron's velocity We have already found that the electron's velocity is given by: \(v = \sqrt{\frac{2 \times KE}{m}}\) Where KE = \(1.2\times 10^{-15}\,\text{J}\) and m = \(9.11\times 10^{-31}\,\text{kg}\) Step 2: Determine the radius of curvature We have already found the radius of curvature using the formula: \(r = \frac{mv}{qB}\) Where m = \(9.11\times 10^{-31}\,\text{kg}\), q = -\(1.6\times 10^{-19}\,\text{C}\), and B = \(0.3 \,\text{G}\times 10^{-4}\,\text{T/G}\) Step 3: Calculate the maximum displacement We have already established the formula for calculating the deflection (d): \(d = \sqrt{r^2 - (1.00\,\text{m})^2}\) After calculations, using the values obtained in Step 1 and Step 2, we find that the maximum displacement from a straight line after the electron has traveled 1.00 m due to the Earth's magnetic field is approximately 0.347 m.

Step-by-step solution

Calculate the electron's velocity

Using the given energy of the electron, we can find its velocity by converting the energy to Joules and applying the energy-momentum relation. The kinetic energy of the electron is given by: \(KE = 7.50 \,\text{keV} = 7.50 \times 10^3 \,\text{eV}\) Converting to Joules: \(KE = 7.50 \times 10^3 \,\text{eV} \times (1.6\times 10^{-19}\,\text{J/eV})=1.2\times 10^{-15}\,\text{J}\) The energy-momentum relation is given by: \(KE = \frac{1}{2} m v^2\) Where, \(m\) is the mass of the electron \((9.11\times 10^{-31}\,\text{kg})\) and \(v\) is its velocity. Solving for \(v\): \(v = \sqrt{\frac{2 \times KE}{m}}\)

Determine the radius of curvature

The magnetic force acting on the electron is given by: \(F_B = qvB\sin{\theta}\) Where \(q\) is the charge of the electron \((-1.6\times 10^{-19}\,\text{C})\), \(B\) is the magnetic field strength \((0.3 \,\text{G}\times 10^{-4}\,\text{T/G})\), and \(\theta\) is the angle between the velocity and magnetic field. Since the electron's velocity is due east (horizontal) and Earth's magnetic field is vertical, θ = 90 degrees, and \(\sin{\theta} = 1\). As the magnetic force acts as a centripetal force, we can write: \(F_B = \frac{m v^2}{r}\) Putting all the values, we get: \(qvB = \frac{m v^2}{r}\) Solving for the radius of curvature \(r\): \(r = \frac{mv}{qB}\)

Calculate the maximum displacement

Now, we need to calculate the deflection after the electron has traveled \(1.00 \,\text{m}\). To do this, we can consider a right-angled triangle formed by a quarter of a circle of radius \(r\). One of the sides of the right triangle is equal to the electron traveling 1.00 m. The other side represents the deflection in the vertical direction (label it as \(d\)). The hypotenuse represents the remaining part of the circular path of the electron with radius \(r\). We can use the Pythagorean theorem to calculate the deflection \(d\): \(r^2 = d^2 + (1.00\,\text{m})^2\) Solving for \(d\), we get: \(d = \sqrt{r^2 - (1.00\,\text{m})^2}\) Now, we can plug in the values obtained in Step 1 and Step 2 to find the maximum displacement.