By the way, I've been stuck on this physics question for a while now.

*When light of wavelength 2.08 * 10^-7 m falls on a photosurface, a voltage of 1.40 V is required to stop the emitted electrons from reaching the anode. What is the largest wavelength of light that will result in the emission of electrons from this photosurface?*

I was thinking since V = E/Q then E = 1.40 * 1.6 * 10-19 which gives E = 2.24 * 10^-19 J, then sub in this value for E = hf_{c} and since h is just planck's constant then you just solve for the critical frequency. Then you use λ = c / f where 'c' is the speed of light and 'f' is the critical frequency, resulting in a wavelength of 8.9 * 10^-7 m, but the answer in my textbook says it's 2.7 * 10^-7 m

halp

Well, we know that the stopping voltage is 1.40V. This corresponds to an energy of 2.24*10^-19 J, as you said. That energy represents the maximum possible kinetic energy that an emitted photoelectron can have, when light of the given wavelength is used.

Now, we use E

_{K max} = hf - W, in order to get the work function. Or, in this case, we use E = hc/λ -W

2.24*10^-19 = (6.63*10^-34) * (3.0*10^8) / (2.08*10^-7) - W.

solving for W gives W = 7.3225*10^-19 J.

Now, we want the cutoff wavelength; the point at which E

_{K max} = 0. (ie. electrons aren't really emitted at all.) Hence, hc/λ - W = 0

W = hc/λ

λ = hc/W

so, λ = (6.63*10^-34) * (3.0*10^8) / (7.3225*10^-19)

which is 2.7*10^-7 m.

The mistake you were making is that the energy you had found was NOT the work function. Rather, it was the amount of energy that was left over once you had subtracted the work function from the energy of the given photons.

If that doesn't make sense I'll try and explain it again; these types of problems were always the most difficult we got in our physics exams. You had to make sure you were visualising what type of energy you were referring to correctly. Talking of which; if you know how to use electron volts as your units rather than joules, it makes it a fair bit simpler.