Question

The wavelength '`lambda`' of a photon and de-Broglie wavelength of an electron have same value. The ratio of energy of a photon to kinetic energy of electron is (m = mass of electron, c = velocity of light, h = Planck's constant) ____________.

Options

• `(2lambda)/"mch"`

• `(lambda"mc")/(2"h")`

• `(lambda"m")/(4"h")`

• `(2lambda"mc")/"h"`

Answer 1

The wavelength '`lambda`' of a photon and de-Broglie wavelength of an electron have same value. The ratio of energy of a photon to kinetic energy of electron is (m = mass of electron, c = velocity of light, h = Planck's constant) `(2lambda"mc")/"h"`.

Explanation:

`lambda_"e" = lambda_"p"`

`therefore "K.E. of photon"/"K.E. of electron" = ("hc"//lambda)/(1/2 "p"^2/"m")`

`(2"mhc")/(lambda"p"^2) = (2"mhc"lambda)/(lambda^2 "p"^2)`

`(2"mc"lambda)/("h"^2) xx "h"`

`= (2"mc"lambda)/"h"`