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

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

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. 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 kinetic energy of an electrons is equal to the energy of photons. Show that de-Broglie wavelength of this electrons is less than the wavelength associated with a photon.

De Broglie wavelength lambda is proportional to

De Broglie wavelength lambda is proportional to

What will be the kinetic energy of a proton if the de Broglie wavelength associated with a proton is 165Å ?