Interactions of Light and Matter

[EskShift]

Homework Statement

The tungsten cathode in an electron microscope has a work function of 4.5 eV and is heated to release electrons that are initially at rest (near the cathode). The electrons are accelerated by a potential difference that creates a beam with a de Broglie wavelength of 0.040 nm.

Calculate the kinetic energy, in Joules, of an electron in the beam.

Homework Equations

Ek = hf - w

F = c / $$\lambda$$

The Attempt at a Solution

Ek = hf - w
F = c / $$\lambda$$
Therefore, Ek = h x (c/$$\lambda$$) - w
Ek = 6.63 x 10^-34 x (3 x 10^8/0.040 x 10^-9) - 4.5
Ek = -4.5?
Feel like I am doing something very wrong here...

 

[Doc Al]

How does the de Broglie wavelength relate to the particle's momentum?

 

[tomcue]

I can provide a response to this content by first clarifying the equations and concepts used in the attempt at a solution. The first equation, Ek = hf - w, is the equation for the kinetic energy of an electron, where h is Planck's constant, f is the frequency of the electron, and w is the work function of the cathode. The second equation, F = c / \lambda, is the equation for the frequency of an electron, where c is the speed of light and \lambda is the de Broglie wavelength.

In the attempt at a solution, the use of the second equation is incorrect. The frequency of an electron is not dependent on the de Broglie wavelength, but rather on the energy of the electron. The correct equation for the frequency of an electron is f = Ek/h.

To calculate the kinetic energy of the electron, we can use the equation Ek = hf - w, where we know the value of h and w. However, we do not have the value of the frequency, so we need to find it using the second equation, f = Ek/h. Rearranging this equation, we get Ek = f x h. We can substitute this value of Ek into the first equation, giving us Ek = (f x h) - w.

To find the frequency of the electron, we can use the equation F = c / \lambda, where we know the value of c and \lambda. Rearranging this equation, we get f = c / \lambda. Substituting this value of f into the previous equation, we get Ek = (c / \lambda) x h - w.

Now, we can plug in the values given in the homework statement. The speed of light, c, is 3 x 10^8 m/s. The de Broglie wavelength, \lambda, is 0.040 nm, which is equal to 0.040 x 10^-9 m. The work function, w, is 4.5 eV, which is equal to 4.5 x 1.6 x 10^-19 J. Plugging these values into the equation, we get:

Ek = (3 x 10^8 / 0.040 x 10^-9) x (6.63 x 10^-34) - (4.5 x 1.6 x 10^-19)
Ek =