An electron of mass m and a photon have same energy E.The ratio of de-Broglie wavelengths associated with them is (C being velocity of light0)
Here C is speed of light

Options

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

Answer

D

Explanation

de-Broglie wavelength of electron `lambda_(e)=(h)/(p)`
But `p=sqrt(2mE)`
`therefore lambda_(e)=(h)/(sqrt(2mE))` ……(1)
`E=hf=(hc)/(lambda_(p))`
`therefore lambda_(p)=(hc)/(E )` ……(2)
`therefore (lambda_(e))/(lambda_(p))=(h)/(sqrt(2mE))xx(E )/(hc)`
`therefore (lambda_(e))/(lambda_(p))=(1)/(c)sqrt(E )/(2m)`