Physics · Dual Nature Of Radiation And Matter · NEET
Yes, but it is a special case. de Broglie said every moving thing has lambda = h/p. For a photon, momentum p = E/c = h*nu/c. Put this in lambda = h/p and you get lambda = hc/E = c/nu. This is just the normal wavelength of the light itself. So for a photon the 'de Broglie wavelength' and the ordinary wavelength are the same number. The photon is a massless particle, so we cannot use the mass formula on it.
A photon has zero rest mass, so its full energy comes from motion and the correct relation is E = pc (this gives p = E/c). An electron has mass m and, at NEET speeds, is non-relativistic, so its kinetic energy is E = (1/2)mv^2 = p^2/(2m) (this gives p = sqrt(2mE)). Different energy-momentum links give different wavelength formulas even for the same energy E.
The photon. Compare lambda_photon = hc/E with lambda_electron = h/sqrt(2mE). Their ratio is lambda_photon/lambda_electron = c*sqrt(2m/E). Because c is very large (3x10^8 m/s), this ratio is much bigger than 1, so the photon wavelength is far longer. A simple way to remember: a photon of energy E carries much more momentum-per-... no; instead recall photon p = E/c is tiny (divide by big c), tiny momentum means big lambda = h/p.
Then their wavelengths are exactly equal. lambda = h/p depends only on momentum, so same p means same lambda for any particle. The photon and electron look different only when you fix the energy, because photon energy is E = pc while electron energy is E = p^2/2m. Always check whether the question fixes energy or momentum.
Write both wavelengths in terms of E. Photon: lambda_ph = hc/E. Electron: lambda_e = h/sqrt(2mE). Divide: lambda_ph/lambda_e = (hc/E) * (sqrt(2mE)/h) = c*sqrt(2mE)/E = c*sqrt(2m/E). If instead the question asks lambda_e/lambda_ph, just flip it: (1/c)*sqrt(E/2m).
An electron of mass m and a photon have the same energy E. The ratio of the de Broglie wavelengths associated with them is (c being the velocity of light):
A photon and an electron (mass m) have the same energy E. The ratio lambda_photon/lambda_electron of their de Broglie wavelengths is: (c is the speed of light)
Electrons of mass m with de Broglie wavelength lambda fall on the target in an X-ray tube. The cutoff wavelength (lambda_0) of the emitted X-ray is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Only if they have the same momentum, because lambda = h/p depends on momentum alone. If instead they have the same energy E, they differ: photon lambda = hc/E and electron lambda = h/sqrt(2mE), and the photon wavelength is much longer.
lambda_photon / lambda_electron = c * sqrt(2m/E), where m is the electron mass and E is the common energy. Flip it for the reverse ratio: lambda_electron / lambda_photon = (1/c) * sqrt(E/2m).
That formula comes from E = p^2/(2m), which needs a rest mass m. A photon is massless, so it uses E = pc instead, giving lambda = hc/E. Putting a mass into the photon formula is the most common NEET trap here.
The electron. Electron p = sqrt(2mE) while photon p = E/c. Because c is huge, dividing E by c makes the photon momentum tiny, so its wavelength (lambda = h/p) is large.
Both. NCERT states lambda = h/(mv), so a larger mass m or a larger speed v gives a smaller wavelength. That is why a cricket ball has an unmeasurably small wavelength but an electron has an X-ray-sized one.
