Physics · Dual Nature Of Radiation And Matter · NEET
Photon: E = hf = hc/lambda, momentum p = h/lambda = E/c. Work function: phi0 = h*f0 = hc/lambda0. Einstein equation: KEmax = hf - phi0 = h(f - f0). Stopping potential: eVs = KEmax = hf - phi0. Max speed: KEmax = (1/2)m*vmax^2. de Broglie: lambda = h/p = h/(mv). de Broglie from energy: lambda = h/sqrt(2mK). de Broglie of accelerated charge: lambda = h/sqrt(2mqV). Electron shortcut: lambda = 12.27/sqrt(V) angstrom. X-ray cutoff (Duane-Hunt): lambda_min = hc/(eV). Photons per second: N = P/(hf) = P*lambda/(hc).
Use h = 6.63 x 10^-34 J s (or 6.6 x 10^-34 when the paper gives it). For energy in eV, the golden shortcut is hc = 1240 eV nm. So threshold wavelength lambda0 (in nm) = 1240 / phi0 (phi0 in eV). This single trick solves most work-function and threshold questions in one line.
Use lambda = h/p when you already know momentum or speed (p = mv). Use lambda = h/sqrt(2mK) when you know kinetic energy K. If a charged particle is accelerated through V volts, then K = qV, so lambda = h/sqrt(2mqV). All three are the same formula written for different given data.
For an electron accelerated through V volts, lambda = 12.27/sqrt(V) angstrom. It is just lambda = h/sqrt(2m_e*e*V) with electron mass and charge already plugged in. It is exam-legal and saves time. Example: V = 81 V gives lambda = 12.27/9 = 1.36 angstrom = 0.136 nm.
For a photon (light): E = hc/lambda and p = E/c = h/lambda. For matter (electron, neutron, proton): lambda = h/p = h/sqrt(2mK). Light: energy first, then p = E/c. Matter: momentum first, then lambda = h/p. Do not use E = pc for matter - that is only for massless photons.
Each photon carries energy E = hf = hc/lambda. If a source gives power P watts, then N = P / E = P*lambda/(hc) photons per second. Keep wavelength in metres and power in watts.
The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of 3.3 x 10^-3 W, will be (h = 6.6 x 10^-34 J s):
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):
The de Broglie wavelength associated with an electron, accelerated by a potential difference of 81 V, is given by:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Einstein's photoelectric equation KEmax = hf - phi0 (with eVs = KEmax) and de Broglie wavelength lambda = h/p. Nearly every year one question comes directly from each of these.
hc = 1240 eV nm. This is the single most useful number in the chapter. Threshold wavelength lambda0 (nm) = 1240 / phi0 (eV), and photon energy E (eV) = 1240 / lambda (nm).
Yes. For an electron accelerated through V volts, lambda = 12.27/sqrt(V) angstrom. It comes from lambda = h/sqrt(2m_e e V) with electron constants substituted.
A photon's de Broglie wavelength equals its actual wavelength, because lambda = h/p and for a photon p = h/lambda. But use E = hc/lambda for photon energy, not lambda = h/sqrt(2mK).
eVs = KEmax = hf - phi0. So the stopping potential Vs measures the maximum kinetic energy of photoelectrons in volts. A graph of Vs versus frequency f has slope h/e.
