Register to reply

Photoelectric Effect ~ threshold frequency dependent on incident light?

Share this thread:
ldesai149
#1
Jul29-11, 01:12 AM
P: 3
1. The problem statement, all variables and given/known data

True/False: In the photoelectric effect, the cut-off (threshold) frequency depends on the intensity of incident light.

2. Relevant equations

hv = W + 1/2mv^2

v = W/h

3. The attempt at a solution

The threshold frequency depends on the wavelength of incident light but not the intensity (??)
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
tiny-tim
#2
Jul29-11, 03:48 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,148
Hi ldesai149!
Quote Quote by ldesai149 View Post
not the intensity (??)
That's right

the whole point of the experiment is that even if you reduce the intensity so that only one photon goes through at a time, that one photon still needs to have the threshold energy (or frequency)
The threshold frequency depends on the wavelength of incident light
no, the threshold frequency is fixed

you might as well say "the threshold frequency depends on the frequency of incident light"

or "the minimum height for joining the army depends on the person's height"
epsilonjon
#3
Jul29-11, 04:45 AM
P: 60
I am a bit unclear on this process too. I never understood why an electron could not absorb the energy from several photons at the same instant and gain the required energy this way, rather than having to absorb from just a single photon at a time.

Thanks :-)

thebiggerbang
#4
Jul29-11, 04:53 AM
P: 70
Photoelectric Effect ~ threshold frequency dependent on incident light?

one can't think about the photoelectric effect in terms of classical physics where the intensity of light is proportional to the square of the amplitude of the wave.

In QM, one deals with light as individual packets of light. Imagine, one packet of light is able to interact with and excite one and only one electron. So, if it's energy ain't high enough (frequency below threshold freq) , it will not be able to eject the electron! However, increase in the intensity of light would just imply more number of incoming photons, and thus more electrons getting excited.
;)
ZapperZ
#5
Jul29-11, 05:15 AM
Emeritus
Sci Advisor
PF Gold
ZapperZ's Avatar
P: 29,238
Quote Quote by epsilonjon View Post
I am a bit unclear on this process too. I never understood why an electron could not absorb the energy from several photons at the same instant and gain the required energy this way, rather than having to absorb from just a single photon at a time.

Thanks :-)
The cross section for such absorption is extremely small (look up multiphoton photoemission). This means that it is of very low probability of happening. By the time the next photon comes along at the right location, there's a very good chance that the electron has decayed back to the lower energy state in the conduction band (the lifetime in the excited state in a metal is of the order of femtoseconds). So using ordinary light, as opposed to high-intensity laser, the intensity is too low to cause multiphoton photoemission. So such photoelectric effect using more than just one photon doesn't normally occur.

Zz.
Pranav-Arora
#6
Jul29-11, 05:42 AM
P: 3,807
Hi Idesa149!!

Did you completely check your equations?
You can write
"hv=W+1/2v^2" as:-

"hv=hv0+1/2mv^2"

Now you see, the threshold frequency(v0) depends on the frequency of incident light.

And as tiny-tim stated, threshold frequency is fixed for a particular metal. That is, if you incident a light with a lower frequency and that too with a large intensity, a single elctron wouldn't be emitted.
Ashu2912
#7
Jul29-11, 05:42 AM
P: 96
(1) Threshhold frequency is the characteristic of the metal, it does not depend on the radiation in any way.
(2) If you mean that why does the photoelectric effect's occurence itself depend on the frequency of the radiation/ incident light and not on the intensity, then it is a correct statement.
(3) This is because the minimum energy needed to knock-out an electron is fixed, i.e. the threshhold energy, corresponding to the threshhold frequency, and one quanta of energy and the energy of one photon of radiation is fixed, depending upon it's frequency only. So, even if you "throw" as many photons of radiation as you can, electrons wont be emitted, as not even one of those photons has sufficient energy to take out an electron.
(4) Intensity is related to the number of such photons, of equal energy. The more the intensity/brightness, the more the number of photons. However, as said before, even if you have a very intense light but of insufficient frequency, photons wont be emitted.
Think of it in this way:
"The battle is one-on-one. So even if you have a large army of weak soldiers against the strongest soldier from the opposite kingdom, the larger army loses. Why, because none of them individually matches the stronger soldier...."


Register to reply

Related Discussions
Effect of altering incident light intensity on absorption General Physics 0
Effect of thickness of atmosphere on the incident light on land General Physics 3
Photoelectric Effect, Threshold Frequency etc. Introductory Physics Homework 1
Threshold frequency in light Introductory Physics Homework 3
Threshold Frequency: PE effect General Physics 5