1.

The energy of a photon is equal to the K.E. of a proton. The energy of a photon is E. If lamda_(1) and lamda_(2)are the de broglie wavelengths of proton and photon, then (lamda_(1))/(lamda_(2)) is proportional to :

Answer»

<P>`E^(@)`
`E^(1//2)`
`E^(-1)`
`E^(-2)`

Solution :For photon `E_(PH)=(hc)/(lambda_(2)), lambda_(2)=(hc)/(E_(ph))`
For photon `E_(p)=1//2 mv^(2)=((mv)^(2))/(2m)=(p^(2))/(2m)`
`rArr p= SQRT(2mE_(p))`
`:.` for proton
`lambda_(1)=(H)/(p)=(h)/(sqrt(2mE_(p)))`
But `E_(ph)=E_(p)=E`
`:. (lambda_(1))/(lambda_(2))-(h)/(sqrt(2mE))xx(E)/(hc)= (sqrt(E))/(sqrt(2)C)`
`:. (lambda_(1))/(lambda_(2)) prop sqrt(3)`


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