Physics · Dual Nature Of Radiation And Matter · NEET
The formula p = mv is only for particles with mass moving slower than light. For a photon the correct relation comes from relativity: E^2 = (pc)^2 + (m0 c^2)^2. Since rest mass m0 = 0, this becomes E = pc, so p = E/c. The photon has momentum because it carries energy and moves at speed c, not because it has mass. This momentum is real and shows up in radiation pressure and the Compton effect.
Energy is E = hf = hc/lambda (units: joule or eV). Momentum is p = E/c = h/lambda (units: kg m/s). Both increase when wavelength decreases, but they are NOT the same quantity. A quick link: p = E/c. So if you know one, divide or multiply by c to get the other. Energy tells you how much a photon can do (like ejecting an electron); momentum tells you the push it can give (radiation pressure).
Both work because c = f x lambda. Energy: E = hf (with frequency) OR E = hc/lambda (with wavelength). Momentum: p = hf/c (with frequency) OR p = h/lambda (with wavelength). Pick whichever matches the data given. If wavelength is given, p = h/lambda is fastest; if frequency is given, use E = hf then p = E/c.
Use the shortcut E(in eV) = 1240 / lambda(in nm). Example: for lambda = 620 nm, E = 1240/620 = 2 eV. This comes from hc = 1240 eV nm. It saves you from plugging in h = 6.63e-34 and dividing by 1.6e-19 every time. Only use it when wavelength is in nanometres.
No. Brightness (intensity) means MORE photons per second, not more energy per photon. Energy per photon depends only on frequency or wavelength (colour), E = hf. A dim blue light has higher energy photons than a bright red light. To get total power you multiply energy per photon by number of photons per second, which leads to the next concept.
If c is the velocity of light in free space, the correct statements about a photon are: A. Energy E = h(nu). B. Velocity of a photon is c. C. Momentum p = h(nu)/c. D. In a photon-electron collision both total energy and total momentum are conserved. E. A photon possesses positive charge.
A photon and an electron (mass m_e), each of 20 eV energy, move in free space. The ratio of the linear momentum of the electron p_e to that of the photon p_ph is: [c = 3e8 m/s, e = 1.6e-19 C, m_e = 9e-31 kg]
The average number of photons per second emitted by a source of monochromatic light of wavelength 600 nm, when it delivers a power of 3.3e-3 W, is: (h = 6.6e-34 J s)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
E = hf = hc/lambda, where h = 6.63e-34 J s, f is frequency, c is speed of light and lambda is wavelength. In electron volts, E(eV) = 1240 / lambda(nm).
p = E/c = hf/c = h/lambda. Its unit is kg m/s. Momentum depends only on wavelength through p = h/lambda.
A photon has zero rest mass and zero charge. It still carries energy and momentum because it always travels at the speed of light c.
From relativity E^2 = (pc)^2 + (m0 c^2)^2. With rest mass m0 = 0 this gives E = pc, so p = E/c. The p = mv rule does not apply to massless particles.
Both decrease as wavelength increases, since E = hc/lambda and p = h/lambda. Shorter wavelength (like X-rays) means higher energy and higher momentum.
