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(2m E)^(1)/(2)`
D. `(1)/(c ) ((2m)/(E))^((1)/(2))`

Answer

A

Explanation

de Broglie wavelength `= (h)/(m v)`
m = mass of particle
v = velocity of particle
h = Planks constant
`E = (1)/(2) m v^(2)`, (Mass of electron = m)
`sqrt((2E)/(m)) = v`
`lambda_(E) = (h)/(msqrt((2E)/(m))) = (h)/(sqrt(2mE))`
Now for photon `E = P c rarr p = (E)/(c)`
`lambda_(p) = (h)/(m v) = (h)/(P) = (h c)/(E)`
`lambda_(E)//lambda_(P) = (h)/(sqrt(2mE)) (E)/(h c) = (1)/(c) sqrt((E)/(2m))`