Photoelectric effect: finding potential difference from frequency and KE

[a.a]

Homework Statement

How would we find the potential difference given the frequency of the incident light and the kenetic energy of an the emitted electrons?

Homework Equations

E= hf-W
V= w/q
e= 1.602*10^-19

The Attempt at a Solution

Given: f= 7.4*10^15 Hz
KE= 1.9eV
Required: potential difference

Solution: KE= 1.9eV * 1.602*10^ -19 J/eV = 3.0438*10^-19

V= 3.0438*10^-19/ 1.602*10^ -19 = 1.9

The answer is supossedd to be 2.7, I don't think I'm doing this the correct way.

 

[rock.freak667]

The stopping potential would be 1.9V, not sure what p.d. they are asking for though.

 

[Redbelly98]

Use the 1st equation you wrote,

E = hf - W

To get W, you need to calculate E and hf.

 

[a.a]

Even when you do that, you wouldn't get 2.7..