Photoelectric effect and a metal plate

[Bindle]

First of all, Hi we haven't even learned the E = ((hc)/λ) in the physics class but we have this question on our test-exam and I hope that it's fairly easy:

Homework Statement

A metal plate is shone upon with a monochromatic light. When the wave-length is 550 nm a photo-electric effect is observed where the fastest photo electrons have the energy of 0.45 eV. When the light have 410 nm wave-length the quickest photo electrons get the energy 1.26 eV. Decide the Planck constant with help of these values.

First λ = 550 x 10^9 m
First E = 0.45 eV
Second λ = 410 x 10^9 m
Second E = 1.26 eV

Homework Equations

E = h*v
λν = c

The Attempt at a Solution

Ok, so I start out with, I'm not sure if I even is to use the conversion of eV to J so I can find the constant but it's the only way I can see that we can do this:

0.45 x (1.60217657 × 10^-19) = E

c/λ= v

(3*10^8)/(550 x 10^9) = v

E/v = h

(0.45 x (1.60217657 × 10^-19))/((3*10^8)/(550 x 10^9)) = h

1.3217957e-16 Js = h

Which doesn't look right. But I guess this has to do with photoelectric effect which haven't learned anything about yet, and that I should use the two values and wavelengths in co-operation in some form to get the Plancks constant.

Anybody who could make me see what to do?

Bindle

 

[rude man]

You need an additional parameter, to wit, the work function of the metal.

Basically, E = hf - W
where W is the work function of the metal, f is the frequency of the impinging light, and E is the maximum observed kinetic energy of the emitted electrons. E cannot be negative.

So you have 2 equations and 2 unknowns: h and W. Solve for h.