Physics · Dual Nature Of Radiation And Matter · NEET
1240 is a ready-made shortcut for h times c. When you put Planck's constant h = 6.63e-34 J s, speed of light c = 3e8 m/s, and divide by the electron charge 1.6e-19 C, the product hc comes out close to 1240 eV nm. So E(eV) = 1240 / lambda(nm) gives photon energy directly in eV when wavelength is in nanometres. This saves you from writing out powers of ten each time.
First make sure the wavelength is in nanometres (nm). If it is given in angstrom, divide by 10 (1 nm = 10 angstrom). If it is in metres, multiply by 1e9. Then apply E(eV) = 1240 / lambda(nm). Example: light of 620 nm has E = 1240 / 620 = 2 eV. That is the whole photon energy in one step.
Stopping potential V0 comes from eV0 = KEmax, and KEmax = E - W. Because e (the charge) appears on both sides when energies are in eV, the number for V0 in volts is the same as KEmax in eV. So if KEmax = 1.2 eV, then V0 = 1.2 V. No extra conversion is needed. This is the biggest time-saver in the chapter.
Convert to joules only when the answer must be in SI units, such as finding photocurrent, number of photons per second, or momentum in kg m/s. For pure comparison questions (does emission happen, which metal emits, what is V0) stay fully in eV. Rule: 1 eV = 1.6e-19 J. Multiply eV by 1.6e-19 to get joules.
Rearrange Einstein's equation: W = E - eV0. Work in eV throughout. Example: light of 400 nm gives E = 1240/400 = 3.1 eV. If V0 = 1.1 V, then eV0 = 1.1 eV, so W = 3.1 - 1.1 = 2.0 eV. The work function of the metal is 2.0 eV.
Then KEmax = E - W would be negative, which is impossible. It means no photoelectrons are emitted at all, so KEmax = 0, current = 0, and stopping potential = 0. Always check E versus W first. Emission happens only when photon energy E is greater than or equal to the work function W.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Use E(eV) = 1240 / lambda(nm) for photon energy, then KEmax = E - W, and V0(volts) = KEmax(eV). These three steps solve almost every eV numerical in the chapter.
It is a rounded value of hc expressed in eV nm. The exact figure is about 1239.8 eV nm, but 1240 is accurate enough for NEET and keeps the arithmetic simple.
Yes. E(eV) = 12400 / lambda(angstrom) is the same shortcut for wavelength in angstrom. Pick one system and stay consistent so you do not mix nm and angstrom.
No. Photon energy depends only on frequency or wavelength, so E = 1240/lambda is unchanged by intensity. Intensity changes only the number of photons, which affects current, not KEmax or stopping potential.
First find KEmax in eV using KEmax = E - W, then multiply by 1.6e-19. For example, KEmax = 1.5 eV = 1.5 times 1.6e-19 = 2.4e-19 J.
