The energy of a photon is equal to the kinetic energy of a
proton. The energy of the photon is E. Let `lambda_1`be the de-Broglie wavelength of the
proton and `lambda_2` be the wavelength of the photon. The ratio `(lambda_1)/(lambda_2)` is proportional to
(a)`E^0` (b) `E^(1//2)` (c ) `E^(-1)` (d)`E^(-2)`

Options

A. `E^0`
B. `E^(1//2)`
C. `E^(-1)`
D. `E^(-2)`

Answer

B

Explanation

Energy of the photon E and it.s wavelength `lamda_2` are related as follows :
`E=(hc)/lamda_2`
`implies lamda_2=(hc)/E" "...(i)`
Further, it is given that energy of photon is equal to kinetic energy of proton.
`E=1/2mv^2`
`m^2v^2=2mEimpliesmv=sqrt(2mE)`
The de Broglie wavelength of proton can be written as follows :
`lamda_1=h/(mv)=h/(sqrt(2mE))" "...(ii)`
From equations (i) and (ii) , we can write the following :
`lamda_1/lamda_2=(h/sqrt(2mE))/((hc)/E)`
`lamda_1/lamda_2=E/(csqrt(2mE))=sqrtE/(csqrt2m)`