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de Broglie Wavelength of Matter Waves


The de Broglie wavelength of a particle of momentum p is given by \begin{align} \lambda=h/p. \end{align} If a particle moves with a relativistic speed then its energy E is related to momentum p by \begin{align} E^2={(pc)^2+(mc^2)^2}, \end{align} where $m$ is the rest mass of the particle and $mc^2$ is its rest mass energy.

Problems from IIT JEE

Problem (IIT JEE 2004): A proton has kinetic energy $E={100}\;\mathrm{keV}$ which is equal to energy of a photon. Let $\lambda_1$ be the de-Broglie wavelength of the proton and $\lambda_2$ be the wavelength of the photon. The ratio $\lambda_1/\lambda_2$ is proportional to

  1. $E^0$
  2. $E^{1/2}$
  3. $E^{-1}$
  4. $E^{-2}$

Solution: The de-Broglie wavelength of a particle of mass $m$, momentum $p$, and kinetic energy $E$ is given by, \begin{align} \lambda_1=h/p=h/\sqrt{2mE}. \end{align} The wavelength of a photon of energy $E$ is given by, \begin{align} \lambda_2=hc/E. \end{align} Divide the first equation by second to get, \begin{align} \lambda_1/\lambda_2=\sqrt{E/(2mc^2)}.\nonumber \end{align} The readers are encouraged to calculate $\lambda_1$ and $\lambda_2$ for given energy.


  1. Photoelectric Effect
  2. de Broglie Wavelength of Matter Waves
  3. Davisson Germer Experiment
JEE Physics Solved Problems in Mechanics