The energy of a photon is E which is equal to the kinetic energy of a proton.If lambda1 be the de-Broglie wavelength of the proton and lambda2 be the wavelength of the photon,then the ratio (lambda_(1))/(lambda_(2)) is proportional to

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)

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

de-Broglie wavelength lambda is

A proton with KE equal to that a photono (E = 100 keV). lambda_(1) is the wavelength of proton and lambda_(2) is the wavelength of photon. Then lambda_(1)/lambda)_(2) is proportional to:

The de Broglie wavelength lambda of a particle

De Broglie wavelength lambda is proportional to

A proton has kinetic energy E = 100 keV which is equal to that of a photon. The wavelength of photon is lamda_(2) and that of proton is lamda_(1) . The ratio of lamda_(2)//lamda_(1) is proportional to

For a proton accelerated through V volts , de Broglie wavelength is given as lambda =