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

Options

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

Answer

A

Explanation

For electron `lamda_e=h/(sqrt(2mE))`
for Photon `E = pc implies lamda_(Ph)=(hc)/E`
`implies lamda_e/(lamda_(Ph))=h/(sqrt(2mE))xxE/(hc) = (E/(2m))^(1//2)1/c`