# 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

- $E^0$
- $E^{1/2}$
- $E^{-1}$
- $E^{-2}$

**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.

## Related

- Photoelectric Effect
- de Broglie Wavelength of Matter Waves
- Davisson Germer Experiment