Physics · Dual Nature Of Radiation And Matter · NEET
No. Work function is the energy to remove an electron from the SURFACE of a solid metal (a collective, bulk property). Ionization energy is the energy to remove an electron from a single ISOLATED atom in gas form. Ionization energy is usually much larger. For example, sodium metal has a work function near 2.75 eV, but the ionization energy of a lone sodium atom is about 5.14 eV.
Both units work, but eV is convenient because single-electron energies in Joules are tiny (about 10 to the power minus 19). One eV = 1.6 x 10^-19 J. In NEET numericals, use the shortcut E(eV) = 1240 / lambda(nm). This gives photon energy directly in eV, so you can compare it to the work function without touching large powers of ten.
No. Work function depends ONLY on the nature and surface of the metal. It does not change with intensity, colour of light, or how long you shine. Intensity changes the NUMBER of photoelectrons, not the energy needed to release one. This is a favourite NTA trap.
They describe the same barrier in two languages. Work function phi is the energy barrier; threshold frequency nu-zero is the minimum light frequency that supplies exactly that energy. They are linked by phi = h x nu-zero. A high work function metal automatically has a high threshold frequency.
Among common metals, Caesium (Cs) has one of the lowest, about 2.14 eV. Low work function means even low-energy visible or near-visible light can eject electrons, so such metals are used in photocells and photocathodes. Platinum and other transition metals have high work functions and need ultraviolet light.
The threshold (longest) wavelength that can still cause emission is lambda-zero = hc / phi. Using the shortcut, lambda-zero(nm) = 1240 / phi(eV). Any wavelength longer than this carries too little energy, so no electrons come out no matter how bright the light.
The work functions of Caesium (Cs), Potassium (K) and Sodium (Na) are 2.14 eV, 2.30 eV and 2.75 eV respectively. If incident electromagnetic radiation has an energy of 2.20 eV, which of these photosensitive surfaces may emit photoelectrons?
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):
For a metal of work function 6.6 eV, which of the following wavelengths of incident radiation does NOT give rise to the photoelectric effect? (Take Planck's constant as 6.6 x 10^-34 J.s)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The SI unit is the Joule (J), but in problems it is almost always given in electron-volts (eV), where 1 eV = 1.6 x 10^-19 J.
It is written as the Greek letter phi (often phi-zero) or as W. NEET uses both; they mean the same quantity.
For NEET, treat it as a fixed property of the metal and its surface. Slight changes with surface condition exist but are ignored at this level.
phi = h x nu-zero, where h is Planck's constant (6.63 x 10^-34 J.s) and nu-zero is the threshold frequency.
No, in the classical photoelectric effect each electron absorbs only one photon. If that single photon's energy is below the work function, no emission occurs.
