Calculating Work Function: Light Sources & Photoelectrons

AI Thread Summary
To calculate the work function of the metal, use the equation K = hf - φ, rewriting it as K = hc/λ - φ for the given light sources. The first light source with wavelength λ yields photoelectrons with a maximum kinetic energy of 1.00 eV, while the second source with half that wavelength produces photoelectrons with 4.00 eV. Set up two separate equations based on these kinetic energies and solve them simultaneously to find the work function φ. This approach is necessary since there are two unknowns to determine. The discussion emphasizes the importance of forming and solving two equations for accurate results.
phy
A light source of wavelength lambda illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1.00 eV. A second light source with half the wavelength of the first ejects photoelectrons with a maximum kinetic energy of 4.00 eV. What is the work functin of the metal?

I know I can use the equation for maximum kinetic energy defined by K=hf - phi where K is the maximum kinetic energy for emmitted electrons, h Planck's constant and f is frequency. Since we don't have frequency, I thought of rewriting the equation as K=hc/lambda - phi but now I don't know how to incorporate both light sources and to actually to solve for phi. Do I have to write two separate linear equations and solve them or is there something esle I'm supposed to be doing? Help please? :confused:
 
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You have the right idea. Two separate equations and solve them. Go with it.
 
phy said:
Do I have to write two separate linear equations and solve them

Bingo! :smile:

Any time you have two unknown quantities, you should suspect that you'll have to set up two equations for them, and solve them together.
 
Haha I just reread the question. I think I was having another blonde moment just a little while ago. Thanks for your help guys :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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