Maximum kinetic energy / cutoff potential problem

  • Thread starter chef99
  • Start date
  • #1
75
4

Homework Statement



Light with a wavelength of 630nm is directed at a metallic surface that has a work function of 1.5 eV. Find:


a) The maximum kinetic energy of the emitted electrons.

b) The cutoff potential required to stop the photoelectrons.


Homework Equations


Ek = hc / λ - W

Ek = eVo

The Attempt at a Solution



a)
λ = 630nm = 6.3 x10-7m

W = 1.5eV (1.60x10-19J/eV)

W = 2.4 x10-19J


Ek = hc / λ - W

Ek = (6.63 x10-34J)(3.0 x108m/s) / (6.3 x10-7m) - 2.4 x10-19J


Ek = 7.57 x10-20J

The maximum kinetic energy is 7.57 x10-20J.


b)

Ek = eVo

7.57 x10-20J = (1.60 x10-19) Vo

Vo = 7.57 x10-20J / (1.60 x10-19)

Vo = 0.47 V

The cutoff is 0.47 V.

I am uncertain if my answer to part a) is correct, I have reviewed all my steps and can't find an error, however, for all of the questions I have done so far, the answer has always been x10-19, whereas my answer here, is 7.57 x10-20J is x10-20. Is this an acceptable answer? Is it possible for it to not always be to the -19J? or did I make an error somewhere in my calculations?
 

Answers and Replies

  • #2
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
10,002
3,149
It looks right. Let me suggest for future reference that, for problems of this sort, you do your calculations in eV. Then all you have to do is convert the photon energy to eV. In this particular problem, the answers to (a) and (b) are identical in number because the max. kinetic energy of the electrons in eV is the same as the cutoff potential in Volts. You can always convert to Joules in the end if needed. :smile:
 

Related Threads on Maximum kinetic energy / cutoff potential problem

Replies
5
Views
3K
  • Last Post
Replies
5
Views
13K
  • Last Post
Replies
13
Views
8K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
1K
Replies
1
Views
2K
  • Last Post
Replies
1
Views
6K
Replies
7
Views
4K
  • Last Post
Replies
4
Views
6K
Replies
1
Views
1K
Top