How Is the Work Function of Silver Determined from Cut-off Wavelength?

[Jas014]

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

Homework Equations

f= c/325nm

The Attempt at a Solution

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

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

 

[mikelepore]

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.

 

[Jas014]

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 I am not sure if i take that number and multiply it by 1.6 * 10 ^ -19?

 

[rock.freak667]

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 I am not sure if i take that number and multiply it by 1.6 * 10 ^ -19?

If you divide $1.6 \times 10^{-19}$ , the resulting number will be in eV. So if you want the units to electron-volts, you divide.