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

A photon and an electron both have wavelength 1 Å . The ratio of energy of photon to that of electron is

A photon and an electron both have wavelength 1 Å . The ratio of energy of photon to that of electron is

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)

A photon and an electron have the same wavelength then, the velocity of photon is

For same energy , find the ratio of lambda _("photon" ) and lambda _("electron") (Here m is mass of electron)

For same energy , find the ratio of lambda _("photon" ) and lambda _("electron") (Here m is mass of electron)

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: