# Homework Help: A photoelectric question

1. Mar 17, 2013

### jiayingsim123

1. The problem statement, all variables and given/known data
Radiation of wavelength 240nm gives rise to a maximum photoelectric current, I. The intensity of the incident radiation is maintained constant and the wavelength is now reduced.
State and explain the effect of this change on
(i) The maximum kinetic energy of the photoelectrons

(ii) The maximum photoelectric current I.

3. The attempt at a solution

I understand that for (i) the maximum KE will increase as the wavelength is reduced causing the photon energy to be larger and the maximum KE is independent of the intensity. However, I don't get (ii). My initial answer is that the maximum photoelectric current, I remains constant as the intensity if unchanged. However, the answer says that there are fewer photons per unit time, so the maximum current is smaller. Does frequency also affect the number of electrons emitted per unit time (since frequency is the number of cycles per unit time) and hence affect the magnitude of the photoelectric current?

I hope you can clear up my confusion! Thanks in advance! :)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 17, 2013

### ehild

What do you think the intensity of light means? :)

ehild

3. Mar 17, 2013

### jiayingsim123

Hi there!
Intensity just means the number of photons contained in the light beam. So the higher the intensity, the more photons are contained. I know clearly that if frequency is kept constant, an increase in intensity will increase the photoelectric current because current is the rate of flow of charge and an increase in intensity will increase the number of electrons emitted from the surface. However, in this case, the wavelength of the light beam is reduced, which means that the frequency is reduced as well. How does this affect the intensity and how does this then, affect the maximum photoelectric current?

I hope you can shed some light on this. Thanks!

4. Mar 17, 2013

### ehild

The intensity of light is the energy transferred in unit time through unit area.

ehild

5. Mar 21, 2018

### Chi Zhang

You forgot that intensity of light is also associated with the frequency by the following relation:

I=nhf/A (n=no. of photons emitted per unit time, h=Planck's constant, f=frequency of light, A=area perpendicular to light's direction of propagation)

In this question, it is mentioned that the wavelength is reduced without the intensity changing. In order to effect this, you have to decrease the rate of emission of photons, by virtue of the formula written about.

A closer examination would require you to analyse the effects the increased maximum kinetic energy has on the photoelectric current (I=nAve). However, if you do the mathematics, you would get the same conclusion.

I hope this helps.