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 De broglie wavelength
`lamda_e = (h)/(sqrt(2mE)) ` , for photon E = pc
de-broglie wavelength , `lamda_(ph) = (hc)/( E)`
` therefore (lamda_e)/(lamda_(ph)) = (h)/(sqrt(2mE)) xx (E )/(hc) = ((E )/(2m))^(1//2) 1/c `