Question

5. The orbiting space shuttle moves around Earth well above 99 percent of the atmosphere, yet it still accumulates an electric charge on its skin due, in part, to the loss of electrons caused by the photoelectric effect with sunlight. Suppose the skin of the shuttle is coated with Ni, which has a relatively large work function φ = 4.87 eV at the temperatures encountered in orbit.
(a) What is the maximum wavelength in the solar spectrum that can result in the emission
of photoelectrons from the shuttle’s skin?
(b) What is the maximum fraction of the total power falling on the shuttle that co

Answer 1

(a) \[eV_{0}=hf-\Phi \]so
\[0=hf-\Phi\]then
\[\Phi=hf\]then
\[ \Phi=(hc)/\lambda\] Finally
\[\lambda=(hc)/\Phi\]

Answer 2

Part (b) in the other hand I don't really know where to start. I found the answer online, but I am misunderstanding the explanation. Problem(25) http://www.physics.utah.edu/~woolf/3740_stone/hw3.pdf

Answer 3

You need to integrate the black body spectrum at the temperature of the surface of the Sun from infinite frequency down to the minimum frequency that will eject electrons, then divide by the total area under the spectrum.

Answer 4

Wow I wouldn't have thought to do that but @Carl_Pham's logic sounds perfect the more I think about what that means in physical terms. Follow what he said, I bet that'll give you the answer you're looking for. Careful to make sure you integrate correctly. Post your integral setup here if you get stuck

Answer 5

Alright I integrated \[U=\int_{0}^{\infty}u(\lambda)d \lambda=8 \pi (kT/hc)^4\int_{0}^{\infty} (x^3)/(e^x -1)=(8\pi^5 k^4/15h^3c^3 )T^4\] just to figure out the example they were talking about. So when you figure out
\[\int_{255}^{\infty} u(\lambda) d\lambda \] over \[\int_{0}^{\infty} u(\lambda) d\lambda \] this is suppose to give you the maximum fraction . Question why would this be the maximum fraction?

Answer 6

I understand now, makes total sense because after initial power, this is what is left from the total spectrum.