Physics · Dual Nature Of Radiation And Matter · NEET
It is a MINIMUM. Threshold frequency nu0 is the smallest frequency that can still eject an electron. Any light with frequency above nu0 works. Light with frequency below nu0 fails completely. Students often mix this up because for wavelength it flips: threshold wavelength lambda0 is a MAXIMUM (the longest wavelength that works), since higher frequency means shorter wavelength.
Each photon carries energy E = h times nu. If nu is below nu0, one photon does not have enough energy to free one electron (its energy is less than the work function W). Making the light brighter only sends MORE photons, but each single photon is still too weak. One weak photon plus another weak photon do not add up on one electron. So the current stays exactly zero. This is a favourite NTA trap: intensity does not help below threshold.
You need a SHORTER wavelength (or equal to lambda0). Shorter wavelength means higher frequency means higher photon energy. Emission happens when lambda is less than or equal to lambda0. If the light has a wavelength longer than lambda0, its photons are too weak and no electron is emitted. So lambda0 = hc/W is the LONGEST wavelength that can just cause emission.
Work function W is the minimum ENERGY (in joule or eV) needed to pull one electron out of the metal surface. Threshold frequency nu0 is the FREQUENCY of light that carries exactly that energy: h times nu0 = W. They describe the same barrier, one as energy and one as frequency. Convert with nu0 = W / h. Both depend only on the metal, not on the light you shine.
It depends only on the METAL. Threshold frequency nu0 = W/h and threshold wavelength lambda0 = hc/W are fixed properties of the metal surface, because W is a property of the metal. Changing the incident light's colour or brightness does NOT change nu0 or lambda0. A metal with a large work function (like tungsten) has a high nu0; a metal with a small work function (like caesium) has a low nu0, so it emits even in visible light.
Energy and wavelength are inversely related: E = hc/lambda. As wavelength gets longer, photon energy gets smaller. So the emission condition E greater than or equal to W becomes lambda less than or equal to hc/W. The equal sign gives the maximum allowed wavelength, lambda0 = hc/W. Any wavelength longer than this is below the energy barrier. There is no shortest-wavelength limit for emission; shorter always works (it just gives more kinetic energy).
The work function of a photosensitive material is 4.0 eV. The longest wavelength of light that can cause photoemission from the substance is (approximately):
Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled?
When a metallic surface is illuminated with radiation of wavelength lambda, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2 lambda, the stopping potential is V/4. The threshold wavelength for the metallic surface is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the minimum frequency of light that can just eject an electron from a metal; below it no photoemission occurs. Formula: nu0 = W / h.
lambda0 = hc / W, where W is the work function, h is Planck's constant and c is the speed of light. A quick shortcut in NEET is lambda0 (in nm) = 1240 / W (with W in eV).
Higher. Since nu0 = W/h, a larger work function needs a higher threshold frequency and a shorter threshold wavelength. Metals like caesium have small W (low nu0) and emit even in visible light.
Emission happens only when nu >= nu0, which is the same as lambda <= lambda0. If the frequency is too low (or wavelength too long), no electron comes out.
Yes, when W is in eV the answer lambda0 comes out in nanometre, because hc = 1240 eV nm. This saves time in numericals. For example W = 4 eV gives lambda0 = 310 nm.
