# Need help with modern physics problem

#### whenwill554

Its due tommorow and its a former AP Physics B free response questions. I appreciate any help. It comes with a diagram which I had to put in a different link.

Diagram:
http://img.photobucket.com/albums/v321/blahblah_1/physics1.jpg

Question:
Electromagnetic radiation is incident on the surface S of a material as shown above. Photoelectrons are emitted from the surface S only for radiation of wavelength 500 nm or less. It is found that for a certain ultraviolet wavelength, which is unknown, a potential Vs of 3 volts is necessary to stop the photoelectrons from reaching the anode A, thus eliminating the photoelectric current.
a. Determine the frequency of the 500 nm radiation.
b. Determine the work function for the material.
c. Determine the energy of the photons associated with the unknown wavelength.
d. Determine the unknown wavelength.

#### physicsquest

PHF Helper
Energy of the photon = work function + k.e. of emitted electron

E = h f where h = planck's constant and f = frequency = c / lamda
where lamda is the wavelength = 500 nm and c = 3 x 10^8 m/sec, the vel of light. This energy is just sufficient to eject the electron and therefore corresponds to the work funtion of the material. When the unknown ultraviolet radiation is incident, the elecron accquires extra K.E. which requires a stopping potential of 3V . Find its K.E. and plug it in the first equation and find the energy of this new photon, and then find its wavelength. Google a bit if you have any difficulty.

whenwill554

#### whenwill554

Thank You so much but by work function does it mean
hf=KE + W where work is the stopping potential of 3v? The 3 volts is throwing me off I dont know how to incorporate it into the problem

#### physicsquest

PHF Helper
No . The energy calculated using E = h c/ 500 nm (convert to metres)
is the work function because this is the energy required to just get out of the material. With the uv light, the electron not only gets out but has excess energy which is used up when it encounters the 3 v potential.
So its k.e. 0.5 mv^2 = work done against this potential. Look up the definition of 'electron volt' (Google) .Once you have this energy, plug it into the firat equation along with the energy calculated using
E = h c/ 500 nm . This sum gives you the energy of the ultra violet radiation.

#### physicsquest

PHF Helper
Use qV = (1/2) m v^2 where q = charge of the electron

Energy of photon = K.E. of electron + Work function

hf = qV + hc/500 x 10^ -8.

f = freq of ultra violet radiation.

#### whenwill554

thank you for your help