1.

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))=

Answer»

`(v)/(C)`
`(v)/(2C)`
`(v)/(3c)`
`(v)/(4c)`

Solution :`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)`


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