An electron of mass `m` and a photon have same energy `E`. The ratio of de - Broglie wavelengths associated with them is :

Options

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

Answer

C

Explanation

For electron of energy E,
de - Broglie wavelength , `lamda_(e) = h/p =(h) /sqrt(2mE)`
For photon of energy , `E = hupsilon=(hc)/lamda_p`
`implies lamda_p =(hc)/E`
`:. lamda_e/lamda_p=h/(sqrt(2mE) )xxE/(hc) =1/c (E/(2m))^(1//2)`