The energy of a photon is equal to the kinetic energy of a proton. If `lamda_(1)` is the de-Broglie wavelength of a proton, `lamda_(2)` the wavelength associated with the proton and if the energy of the photon is E, then `(lamda_(1)//lamda_(2))` is proportional to

Options

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

Answer

A

Explanation

Total energy/s=1000 J
Energy released/fission=200MeV
`=200xx1.6xx10^(-13)J=3.2xx10^(-11)J`
`therefore`Number of fission/s=`(1000)/(3.2xx10^(-11))=3.12xx10^(13)`.