The photoelectric effect is a concern in space, where parts of a spacecraft exposed to the sunlight will develop a net positive charge, which can interfere with or damage sensitive cir The panels that cover the exterior of the Space Shuttle are composed of graphite fibers, which have a threshold frequency of 1.2 x10^15 Hz.
A. What is the K.E of an electron ejected from a panel, in J, when a photon having a wavelength of 190 mm is absorbed?
For this part, I tried using Ek= hc/ lambda and got 1.045x10^-18 J. But the answer key is 2.5x10^-19 J.
B. What is the velocity of ejected electron in m/s?
I know I have to use Ek=1/2 mv^2. Correct me if I'm wrong.
C. What is the de Broglie wavelength of this ejected electron in m?
De Broglie wavelength is lambda= h/mv
a. find the energy of the 190nm photon, you have the right formula. Now subtract the threshold energy (E=h f). With the remainder, you have ke.
b. correct, use E=1/2 m v^2
c. correct formula...
A.
K.E. = hc/w - hf
B
Yes.
C.
Yes.
Thank you so much!
To answer part A of the question, you correctly used the equation Ek = hc/λ to find the kinetic energy of the ejected electron. However, it seems that there might be a mistake in the calculation. Let's go through the calculation step by step to see where the discrepancy might be.
First, let's convert the given wavelength from millimeters (mm) to meters (m). Since 1 mm equals 1 x 10^-3 m, the wavelength is 190 mm = 190 x 10^-3 m = 0.19 m.
Now, we can substitute the values into the equation Ek = hc/λ. Recall that Planck's constant (h) is approximately 6.626 x 10^-34 Js, and the speed of light (c) is about 3 x 10^8 m/s.
Ek = (6.626 x 10^-34 Js) * (3 x 10^8 m/s) / 0.19 m
Ek = (1.9878 x 10^-25 Jm) / 0.19 m
Ek ≈ 1.045 x 10^-24 J
It appears that there was an error in your calculation, and the correct value for the kinetic energy is approximately 1.045 x 10^-24 J. However, this value does not match the answer key provided, so it is possible that there is an error or discrepancy between different sources.
Moving on to part B, you are correct in using the equation Ek = 1/2 mv² to calculate the velocity of the ejected electron. We can rearrange this equation to solve for v:
v = √(2Ek/m)
Substituting the known values, Ek ≈ 1.045 x 10^-24 J and the mass of an electron (m) is approximately 9.11 x 10^-31 kg:
v = √(2 * 1.045 x 10^-24 J / 9.11 x 10^-31 kg)
v ≈ 1.549 x 10^6 m/s
Thus, the velocity of the ejected electron is approximately 1.549 x 10^6 m/s.
Finally, for part C, we can use the de Broglie wavelength equation λ = h/mv to find the wavelength of the ejected electron. Substituting the values:
λ = (6.626 x 10^-34 Js) / (9.11 x 10^-31 kg) * (1.549 x 10^6 m/s)
λ ≈ 8.590 x 10^-10 m
Therefore, the de Broglie wavelength of the ejected electron is approximately 8.590 x 10^-10 m.