Question

Consider a neutron (mass $m$) of kinetic energy $E$ and a photon of the same energy. Let $\lambda_n$ and $\lambda_p$ be the de Broglie wavelength of neutron and the wavelength of photon respectively. Obtain an expression for $\dfrac{\lambda_n{\lambda_p}$.}

Hint:

For particles use de Broglie relation $\lambda = \frac{h}{p}$ and for photons use $E = \frac{hc}{\lambda}$. Combining both relations helps compare wavelengths of matter waves and light.

Answer 1

Step 1: De Broglie wavelength of neutron.
The de Broglie wavelength of a particle is given by \[ \lambda = \frac{h}{p} \] where $h$ is Planck’s constant and $p$ is momentum.
For a neutron having kinetic energy $E$, the relation between kinetic energy and momentum is \[ E = \frac{p^2}{2m} \] Therefore \[ p = \sqrt{2mE} \] Hence the de Broglie wavelength of neutron is \[ \lambda_n = \frac{h}{\sqrt{2mE}} \]

Step 2: Wavelength of photon.
For a photon, energy is related to wavelength by the relation \[ E = \frac{hc}{\lambda_p} \] Therefore \[ \lambda_p = \frac{hc}{E} \]

Step 3: Obtain the ratio $\dfrac{\lambda_n}{\lambda_p}$.
Substituting the expressions of $\lambda_n$ and $\lambda_p$ \[ \frac{\lambda_n}{\lambda_p} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}} \] \[ = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc} \] \[ = \frac{E}{c\sqrt{2mE}} \] \[ = \frac{\sqrt{E}}{c\sqrt{2m}} \]

Step 4: Final expression.
Thus the required ratio is \[ \boxed{\frac{\lambda_n}{\lambda_p} = \frac{\sqrt{E}}{c\sqrt{2m}}} \]