the energy of photon is equal to the kinetic energy of a porton,If `lambda_1` is the De-Broglie wavelength of a proton,`lambda_2` the wavelength associated with the proton and if the energt of the protonis E, then `(lamda_1//lambda_2)` is proportional to

Options

A. `E^4`
B. `E^(1//2)`
C. `E^2`
D. E

Answer

b

Explanation

For De-Broglie wavelength of photon ,`E=hv=P^2/(2m)`
`implies p=sqrt(2mE`….(1)
De-Broglie wavelength of photon ,`lambda_1=h/p` …..(2)
De-Broglie wavelength of associate with photon,
`lambda_2=(hc)/(E)
lambda_1/lambda_2=(p)/((hc)/E)=E/sqrt(2mE.c)=sqrt(E)/sqrt(2mE.c)
implies lambda_1/lambda_2alpha sqrtE
therefore lambda_1:lambda_2=E^(1//2)`