An electron(of mass m) and a photon have some 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=speed of light in vacuum)

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$\frac{1}{c} {(\frac{E}{2 m})}^{\frac{1}{2}}$

$\frac{1}{c} {(\frac{2 E}{m})}^{\frac{1}{2}}$

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

$c {(2 m E)}^{\frac{1}{2}}$

answer is A.

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Detailed Solution

$d e - B r o g l i e w a v e l e n g t h f o r e l e c t r o n = λ_{d}$ $= \frac{h}{\sqrt{2 m E}}$

$W a v e l e n g t h o f p h o t o n = λ =$ $\frac{h c}{E}$
Ratio $\frac{λ_{d}}{λ} = \frac{h}{\sqrt{2 m E}} \times \frac{E}{h c} = \frac{1}{c} \sqrt{\frac{E}{2 m}}$

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