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 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 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 photon and an electron both have wavelength 1 Å . The ratio of energy of photon to that of electron is
A proton has kinetic energy E = 100 keV which is equal to that of a photo is lambda_(1) . The ratio of k_(2) // lambda_(1) is proportional to
For same energy , find the ratio of lambda _("photon" ) and lambda _("electron") (Here m is mass of electron)
An electron (e,m) and photon have same energy E. Then the ratio lambda_e : lambda_p is?
Photon and electron are given same energy (10^(-20) J) . Wavelength associated with photon and electron are lambda_(ph) and lambda_(el) then correct statement will be
A photon and electron have got same de-Broglie wavelength. Total energy of an electron is............than that of photon.
The wavelength lambda of a photon and the de-Broglie wavelength of an electron have the same value. Show that the energy of the photon is (2 lambda mc)/h times the kinetic energy of the electron, Where m,c and h have their usual meanings.
AAKASH INSTITUTE-ALTERNATING CURRENT -Assignment (Section-J) (Aakash Chailengers Questions)
