An electron and a photon possess the same de-Broglie wavelength. If `E_(e)` and `E_(ph)` are respectively the energies of electron and photon and v and c are their respective velocities, then `(E_(e))/(E_(ph))=`

Options

A. `(v)/(c)`
B. `(v)/(2c)`
C. `(v)/(3c)`
D. `(v)/(4c)`

Answer

B

Explanation

`lambda=(h)/(sqrt(2mE_(e)))`
Also `lambda=(hc)/(E_(p)), But E_(e)=(1)/(2) mv^(2)`
So `m=(2E_(e))/(v^(2))`
`rArr 2E_(e)m=m^(2)v^(2)=(m^(2)v^(2))=(mv)^(2)=p^(2) [ :. P=((E_(p))/(c))]`
`:.e[(2E_(e))/(v^(2))]E_(e)=(E_(p)^(2))/(c^(2))`
Solving `(E_(e))/(E_(p))=(v)/(2c)`