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 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. 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 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

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

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

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

A photon and an electron have equal energy E . lambda_("photon")//lambda_("electron") is proportional to

A photon and an electron have equal energy E . lambda_("photon")//lambda_("electron") is proportional to