# Threshold Frequency

1. Jun 8, 2008

### DMac

1. The problem statement, all variables and given/known data
Find the threshold frequency for a calcium surface whose work function is 3.33 eV.

2. Relevant equations
Ek = hf - W

3. The attempt at a solution
My textbook doesn't have any examples on this, but I reasoned that threshold frequency would mean that the kinetic energy of the photoelectron is zero.

So,
W = hf
f = W / h

Since the work is given in eV,
W = 3.33 eV x 1.6x10^-16 J/eV
W = 5.34 x 10^-19 J

Since h = 6.63 x 10^-34,

W = hf
W = 8.05 x 10^14 Hz.

However, my textbook says 5.02 x 10^14 Hz. Where did I go wrong?

2. Jun 8, 2008

### LHC

Hm, for some reason I got the same number as you. Maybe wait for someone with more knowledge on this topic to reply?

3. Jun 8, 2008

### rock.freak667

But if you use W=hf and use their answer (5.02x10^14 Hz),you'll get the work function is 2.08eV, which isn't the work function of calcium.

4. Jun 8, 2008

### Redbelly98

Staff Emeritus
Weird, I get the same thing you do.

5. Jun 8, 2008

6. Jun 8, 2008

### DMac

Thank you so much, guys. I really appreciate your help. =D

7. Jun 8, 2008

### Ed Aboud

No problem DMac.

8. Jun 8, 2008

### a.a

Can someone please explain why we have to convert the Work function into jules?

9. Jun 8, 2008

### rock.freak667

because f=W/h

the units of W is J and the units of h is Js

so the units of f would be $\frac{J}{Js}=s^{-1}=Hz$

if you didn't want to convert from eV to J, you'd need to convert the units of h to eVs