An electron (of mass m) and a photon have the same energy E in the range of 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 vaccum)

Options

A. `(1)/(c)((E )/(2m))^((1)/(2))`
B. `c((E )/(2m))^((1)/(2))`
C. `sqrt(2mE)/(c )`
D. `((E )/(2m))^((1)/(2))`

Answer

A

Explanation

de-Broglie wavelength of electron `lambda_(c)=(h)/(sqrt(2mE))`
de-Broglie wavelength of photon
`lambda_(p)=(hc)/(E )[because E =(hc)/(lambda)]`
`therefore` Ratio `(lambda_(c))/(lambda_(p))=(h)/(sqrt(2mE))xx(E )/(hc)`
`=(1)/(2)(sqrt(E )/(2m))=(1)/(c )((E )/(2m))^((1)/(2))`