Solve Problem Set 3: Electrons, Magnets, Radiation & Photons

[Shackleford]

4. An electron entering Thomson's e/m appartus has an initial velocity of 0.50 x 10^7 m/s. Lying around the lab is a permanent horseshoe magnet of strength 1.3 x 10^-2 T, which you would like to use. (a) What electric field will you need in order to produce zero deflection of the electrons as they travel through the apparatus? (b) The length of nonzero E and B fields is 2.0 cm. When the magnetic field is turned off, but the same electric field remains, how far in the vertical direction will the electron beam be deflected over this length?

For (b), I'm a bit uncertain in using length as the adjacent side in using tangent.

20. (a) At what wavelength will the human body radiate the maximum radiation? (b) Estimate the total power radiated by a person of medium build (assume an area given by a cylinder of 165-cm height and 13-cm radius).

34. What is the threshold frequency for the photoelectric effect on lithium (W = 2.93 eV)? What is the stopping potential if the wavelength of the incident light is 400 nm?

48. If a 6.0-keV photon scatters from a free proton at rest, what is the change in the photon's wavelength if the photon recoils at 90 degrees?

I'm a bit uncertain about this one since I didn't use the 6.0-keV in equation.

If anyone wants to do a quick check of the set, I would appreciate it. Thanks.

 

[Jhero]

Sure, I would be happy to review your answers.

4. (a) To produce zero deflection, the electric field must be equal and opposite to the magnetic force on the electron. The magnetic force is given by F = qvB, where q is the charge of the electron, v is its velocity, and B is the magnetic field strength. In this case, the magnetic force must be equal and opposite to the electric force, which is given by F = Eq, where E is the electric field and q is the charge of the electron. Setting these two forces equal to each other and solving for E, we get E = vB = (0.50 x 10^7 m/s)(1.3 x 10^-2 T) = 6.5 x 10^5 V/m.

(b) The force on the electron due to the electric field is given by F = Eq, where E is the electric field and q is the charge of the electron. Since the electron is moving perpendicular to the electric field, the only component of the force that will cause deflection is in the vertical direction. Therefore, we can use trigonometry to calculate the vertical displacement of the electron over a distance of 2.0 cm. Using the tangent function, we have tanθ = F/F_vert, where θ is the angle of deflection, F is the total force on the electron, and F_vert is the vertical component of the force. Solving for F_vert, we get F_vert = Ftanθ = (Eq)tanθ. Plugging in the values for E and q, we get F_vert = (6.5 x 10^5 V/m)(1.6 x 10^-19 C)tanθ. The angle of deflection, θ, can be calculated using the length of the nonzero E and B fields and the initial velocity of the electron. Using the formula θ = vL/B, where L is the length of the nonzero fields, we get θ = (0.50 x 10^7 m/s)(2.0 cm)/(1.3 x 10^-2 T) = 7.7 x 10^-6 rad. Plugging this value into our equation for F_vert, we get F_vert = (6.5 x 10^5 V/m)(1.6 x 10^-19 C)(7.7 x 10^-