What is the minimum frequency of light needed to eject electrons from a metal?

[supermenscher]

What minimum frequency of light is needed to eject electrons from a metal whose work function is 4.1*10^-19J.

I converted the work function to electron volts and I know that hf = KE+W
and f=KE+W/h but where to I go from there?

 

[quantumdude]

Good start. The minimum frequency corresponds to KE=0. That is, the photon is just energetic enough to get the electron out of the metal, but not to give it any KE.

 

[supermenscher]

So is it just f=W/h I did that and got the wrong answer...and can you help me with another problem.

X-rays of wavelength 0.120nm are scattered from a carbon block. What is the compton wavelength shift for photons detected at the 45 and 180 degrees relative to the incident bean. I did wavelength` = wavelength +h/mc (1-cos theta) I did the problem both ways, solving for wavelength` and wavelength and got the answer wrong both times...any help?

 

[Doc Al]

So is it just f=W/h I did that and got the wrong answer...and can you help me with another problem.

If you got the wrong answer, then you must have done something wrong. Show your work and let's find out.

X-rays of wavelength 0.120nm are scattered from a carbon block. What is the compton wavelength shift for photons detected at the 45 and 180 degrees relative to the incident bean. I did wavelength` = wavelength +h/mc (1-cos theta) I did the problem both ways, solving for wavelength` and wavelength and got the answer wrong both times...any help?

Again, your methods seem OK to me. Show us the details of what you did.

 

[supermenscher]

For the first one I did f=W/h = 4.1*10^-19J (1/1.6*10^-19J)=2.565ev
f=2.565ev/6.626*10^-34=3.86710^33

For the second one, which wavelength do i solve for, wavelength` or wavelength because it says with respect to the incident beam

 

[Doc Al]

For the first one I did f=W/h = 4.1*10^-19J (1/1.6*10^-19J)=2.565ev
f=2.565ev/6.626*10^-34=3.86710^33

For some reason, you converted the work function from J to ev. Don't. The value of h that you used has units of Joule-sec.

For the second one, which wavelength do i solve for, wavelength` or wavelength because it says with respect to the incident beam

I will rewrite the Compton formula like this:
\lambda_f = \lambda_i + \frac{h}{m_e c} (1-cos\theta)

You are given the initial wavelength of the incident beam \lambda_i; solve for the final wavelength \lambda_f.