Photoelectric Effect Homework: Calc Max KE & Stopping Potential

[ishterz]

Homework Statement

Em radiation of frequency .88X10^-15 Hz falls upon a surface of work function 2.5 eV.

A)Calculate Max KE

B) If a nearby electrode is made negative with respect to the first surface using a pd V, what value is required for V if it is to be just sufficient to stop any of the photoelectrons from reaching the negative electrode?


2. The attempt at a solution

I got part A as I did Max KE = hf + work function energy in joules = 9.81x10^-19

I'm not too sure about part b, is the stopping potential just 2.5 eV then?


Thank you :)

 

[ideasrule]

I got part A as I did Max KE = hf + work function energy in joules = 9.81x10^-19

It should be hf - work function, because the ejected electrons have to climb up the potential wells of the atoms they were bound to, losing energy in the process. Also, are you sure the frequency is 0.88E-15 Hz and not 0.88E15 Hz?

I'm not too sure about part b, is the stopping potential just 2.5 eV then?

No. The stopping potential is the potential difference that the ejected electrons can go through before losing all of its kinetic energy. So, what potential difference corresponds to a potential energy difference of Max KE?

 

[Redbelly98]

I got part A as I did Max KE = hf + work function energy in joules = 9.81x10^-19

See ideasrule's note about this calculation. By the way, for energies this small it is common practice to use eV for the energy units.

Moderator's note: thread moved to Introductory Physics