Light of intensity 2.0 W/ m^2 is incident on an aluminim surface. The wavelength of the light is 160nm. For aluminium the work function is 4.2eV.(adsbygoogle = window.adsbygoogle || []).push({});

Find the kinetic energy of the slowest moving electrons

So then Kmin would be the wavelength at which the cutoff frequency occurs, yes?

e = hf -> e = hc / lambda and E = 4.2eV = hc / lambda then find the cutoff wavelength?

At this point im stumped still... please do help thank you in advance!

Heres another one part b of this question

Fidn the average rate per unit area at which photons strike the aluminium surface

P = E / t = hf / t = hc / lambda t and lambda = 160nm.

I = P / A = hc / lambda t A. I know intensity, h ,c , lambda and all i need is something over t A giving units of per second per sq m

Am i right? input would be greatly appreciated!

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Question on Photoelectric effect

**Physics Forums | Science Articles, Homework Help, Discussion**