Question

An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is $(c=speedoflightinvacuum)$

Answer 1

Step1: Given data

1. The mass of the electron is m.
2. The energy of an electron and proton is nearly equal.

Step2: de-Broglie wavelength electron

1. The de-Broglie wavelength of an electron is defined by the form, ${\lambda }_e=\frac{h}{mv}$, where, h is the Planck constant, m is the mass of the electron and v is the velocity of the electron.
2. It is defined by another for, ${\lambda }_e=\frac{h}{\sqrt{2mE}}$, E is the energy of the electron.

Step3: The wavelength of a photon

1. The wavelength of a photon is defined by the form, ${\lambda }_p=\frac{hc}{E}$, where, E is the energy of the photon and c is the velocity of light in free space.

Step4: Finding the ratio

As we know,

The de-Broglie wavelength of an electron is ${\lambda }_e=\frac{h}{\sqrt{2mE}}$

And the wavelength of a photon is ${\lambda }_p=\frac{hc}{E}$

So,

$$\begin{gathered} \frac{{\lambda }_e}{{\lambda }_p}=\frac{{\displaystyle \frac{h}{\sqrt{2mE}}}}{{\displaystyle \frac{hc}{E}}}=\frac{1}{c}\sqrt{\frac{E}{2m}} \\ or{\lambda }_e:{\lambda }_p=\frac{1}{c}\sqrt{\frac{E}{2m}}:1 \end{gathered}$$

Therefore, the ratio of the de-Broglie wavelength of an electron and the wavelength of a photon is $\frac{1}{c}\sqrt{\frac{E}{2m}}:1$