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Solving for Work Function

  • Thread starter Jas014
  • Start date
3
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1. Homework Statement
The wavelength associated with the cut-off frequency, for silver is 325 nm.

a) What is the work function for silver?
b) Find the maximum kinetic energy and speed of the electrons ejected from a silver surface by ultraviolet light of wavelength 254nm


2. Homework Equations

f= c/325nm


3. The Attempt at a Solution

f = c/3.25 * 10 ^ -7 = 9.23 * 10 ^ 14

9.23 * 10 ^14 = hc/325e - work function
 

Answers and Replies

536
2
Try writing the equations completely before you begin to substitute any numbers. The photon energy is: E = hf = hc/lambda = work function + KE_max. The work function is determined by setting KE_max to zero and using the value of your cutoff frequency or wavelength. For part (a), the work function is your only unknown. Once you get it, you can go to part (b) and change the value of the wavelength.
 
Last edited:
3
0
I went

lamda = hc / work function
workfuntion = hc/lamda = 6.626 * 10 ^-34 ( 3 * 10 ^8) / 3.25 * 10 ^ -7 = 6. 11 * 10 - 19

Now im not sure if i take that number and multiply it by 1.6 * 10 ^ -19?
 
rock.freak667
Homework Helper
6,230
31
I went

lamda = hc / work function
workfuntion = hc/lamda = 6.626 * 10 ^-34 ( 3 * 10 ^8) / 3.25 * 10 ^ -7 = 6. 11 * 10 - 19 J

Now im not sure if i take that number and multiply it by 1.6 * 10 ^ -19?
If you divide [itex]1.6 \times 10^{-19}[/itex] , the resulting number will be in eV. So if you want the units to electron-volts, you divide.
 
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