Question

19. Electrons are ejected from a metallic surface with speeds of up to 4.60 X 10^5 m/s when light with a wavelength of 625 nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface?

Answer 1

@wolfe8

Answer 2

I believe we use E=K-W K=1/2mv^2 and E=hf

Answer 3

f=v/wavelength

Answer 4

For part a; i managed to get that.
\[\large K_{max}= hf-\phi\]

Answer 5

Where phi is the work function.
The cutoff frequency is defined as
\[\large \lambda_c= {hf \over \phi}\]
But I can't get the right numerical value.

Answer 6

Yea I was supposed to put + instead of - there. That looks right. What do you mean right numerical value?

Answer 7

oops
\[hc\over \phi\]

Answer 8

oh well, according to my book, phi=1.38 eV which i got. But i can't seem to get the cutoff frequency of lambda_c=3.34 X10^14Hz

Answer 9

Hmmmmmm

Answer 10

Ah we used the wrong Planck's constant. Because our energy for phi is in eV, we have to use 4.13×10^(−15) eV/s for h.

Answer 11

I meant eVs

Answer 12

hmm, nope
phi = 1.38 eV
so the calculation would be
4.13 X10^-15 *.X10^8 / 1.38

Answer 13

that doesn't give me 3.34X10^14

Answer 14

Ok. just use Einstiens equation

since speed is given, u know what is the k.e energy of the electron

and input energy can be found out from wavelength

Einstien says

Input energy = work function + K.E
tada.. :D

Answer 15

is it 1.39 eV ?

Answer 16

@mashy
phi is 1.38 eV; I'm looking for the frequency which is 3.34 X10^14 Hz
I CAN'T get 3.34 X10^14 Hz and I don't know how to get it.

Answer 17

work function = h (threshold frequency) [S.I. units though]

Answer 18

i get exactly that value 3.34 * 10^14

Answer 19

i think u forgot to convert from eV to Joules..

work function = 1.38 * 1.6* 10^-19

Answer 20

Cut off frequency should be as follows:\[\phi=hf _{c}\]\[f _{c}=\phi/h\]Using that and the Planck's constant in EV-s (4.136e-15), I get a cutoff frequency of 3.34e14 (898.4nm).

Answer 21

How did @PsiSquared derive the equation for the cutoff frequency?
@douglaswinslowcooper
@wolfe8
@ybarrap

Answer 22

There is a minimum amount of energy needed to escape,

Emin = h f, and phi is the same as Emin.

Answer 23

derived the cutoff frequency just as @douglaswinslowcooper suggested.

Answer 24

could you elaborate @PsiSquared

Answer 25

Given:\[hf=\phi+E _{K}\]you want to find the frequency of light required to just liberate an electron without giving it any additional energy. The only reason that electron's liberated by the photoelectric effect have kinetic energy is because they're given energy beyond what is needed to free them. If the kinetic energy is zero, then the above equation becomes\[hf=\phi \]This equation describes the energy needed to counter the energy binding the electron to the surface. An electron with that energy would be free but have no kinetic energy.

The work function, φ, tells just tells us the minimum energy required to free said electron. Does that make sense?

Answer 26

yup; thanks!