You are Here: Home >< Physics

# photoelectric effect watch

1. This is to do with the topic of photoelectric effect in AS Physics. I'm struggling on understanding stopping voltage and how maximum kinetic energy of the photo electrons=eVs.

When you shine light on a metal surface, the electrons gain energy. If this energy equals the work function of the surface, only photoelectric emission occurs. This means the electrons only leave the surface. If the electrons gain more than the work function of the surface, photoelectric emission occurs since the electrons leave the surface, but also the electrons gain some kinetic energy. So what does the stopping voltage have to do with this and what exactly is it?

Thanks so much!
2. If a charge q moves through a potential difference V it gains or loses energy Vq
In the common case where the charge accelerates in the field, we say it gains kinetic energy from the field and ½mv² = Vq
where v is the speed the charge reaches. (It started from rest)
The same happens in reverse. If a charge is moving in a field with speed v and has kinetic energy then if the field is such that it opposes the motion (eg a negative charge moving towards the negative plate) then the work done on the charge by the field is Vq and this is equal to the ke removed from the charge. So Vq = ½mv² applies to either accelerating or decelerating a charge in a field. In the photoelectric effect, you use the applied field to stop the electron (you make the plate it's heading for negative) so the field takes the ke away from the electron (decelerates it) rather than gives ke to it.
In this experiment this is useful because by measuring the stopping potential you can work out what the ke of the electron was before it was decelerated by simply applyiny Vse = ½mv²
3. (Original post by Stonebridge)
If a charge q moves through a potential difference V it gains or loses energy Vq
In the common case where the charge accelerates in the field, we say it gains kinetic energy from the field and ½mv² = Vq
where v is the speed the charge reaches. (It started from rest)
The same happens in reverse. If a charge is moving in a field with speed v and has kinetic energy then if the field is such that it opposes the motion (eg a negative charge moving towards the negative plate) then the work done on the charge by the field is Vq and this is equal to the ke removed from the charge. So Vq = ½mv² applies to either accelerating or decelerating a charge in a field. In the photoelectric effect, you use the applied field to stop the electron (you make the plate it's heading for negative) so the field takes the ke away from the electron (decelerates it) rather than gives ke to it.
In this experiment this is useful because by measuring the stopping potential you can work out what the ke of the electron was before it was decelerated by simply applyiny Vse = ½mv²
Thanks a lot!

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 30, 2013
Today on TSR

### Tuition fees under review

Would you pay less for a humanities degree?

### Can I get A*s if I start studying now?

Discussions on TSR

• Latest
Poll
Discussions on TSR

• Latest

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE