1.

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)

Answer»

`(1)/(c)((E )/(2M))^((1)/(2))`
`c((E )/(2m))^((1)/(2))`
`sqrt(2ME)/(c )`
`((E )/(2m))^((1)/(2))`

Solution :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))`


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