Question

An electron (of mass $\mathrm{m}$) and a photon have the same energy $\mathrm{E}$ in the range of few $\mathrm{eV}$. The ratio of the de Broglie wavelength associated with the electron and the wavelength of the photon is. ( $\mathrm{c}=$ speed of light in vacuum)

• A. ${\left(\frac{E}{2m}\right)}^{\frac{1}{2}}$
• B. $\frac{1}{c}{\left(\frac{2E}{M}\right)}^{\frac{1}{2}}$
• C. $C{\left(2mE\right)}^{\frac{1}{2}}$
• D. $\frac{1}{C}{\left(\frac{E}{2m}\right)}^{\frac{1}{2}}$

Answer 1

The correct option is D

$\frac{1}{C}{\left(\frac{E}{2m}\right)}^{\frac{1}{2}}$

The above question can be solved as:

Step 1:

For electron,

De Broglie’s wavelength $$\begin{gathered} \lambda =\frac{h}{p} \\ =\frac{h}{mv} \end{gathered}$$

Step 2:

$\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2$

$=$ $\frac{{\left(\mathrm{mv}\right)}^2}{2m}$

$=$$\frac{{\mathrm{p}}^2}{2\mathrm{m}}$

Hence, $\mathrm{p}=\sqrt{2\mathrm{mE}}$

Step 3:

So, ${\mathrm{\lambda }}_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}}$--------------------------1

Step 4:

For proton,

Energy ⇒ $E=\frac{\mathrm{hc}}{\mathrm{\lambda }}$

So, ${\mathrm{\lambda }}_{\mathrm{p}}=\frac{\mathrm{hc}}{\mathrm{E}}$-----------------------------------2​

Comparing 1 and 2,

$\frac{{\mathrm{\lambda }}_{\mathrm{e}}}{{\mathrm{\lambda }}_{\mathrm{p}}}=\frac{hE}{\sqrt{2mE}hc}$

$=\frac{1}{\mathrm{c}}{\left(\frac{\mathrm{E}}{2\mathrm{m}}\right)}^{\frac{1}{2}}$

So, option (D) is correct.