(1) The energy of a photon is 20eV. Its momentum in kg m/s is
(a) 2.56×10^-27 (b) 5.33×10^-27 (c) 1.066×10^-26 (d) 2.13×10^-26 (e) 3.18×10^-26
Since E=mc^2, momentum of the photon, p = E/c = (20×1.6×10^-19)/(3×10^8). Note that we have converted the energy in eV into joule. The answer is 1.066×10^-26 kg m/s given in option (c).
(2) The wave length associated with an electron having kinetic energy 6eV is
(a) 9A.U. (b) 5A.U. (c) 2.5A.U. (d) 1.5A.U. (e) 0.5A.U.
In the case of electrons, de Broglie wave length, λ = √ [ 150/V] A.U. where V is the accelerating voltage for the electron (= 6 volt since the energy is 6 eV).
Therefore, λ = √25 = 5 A.U.
(a) 2.56×10^-27 (b) 5.33×10^-27 (c) 1.066×10^-26 (d) 2.13×10^-26 (e) 3.18×10^-26
Since E=mc^2, momentum of the photon, p = E/c = (20×1.6×10^-19)/(3×10^8). Note that we have converted the energy in eV into joule. The answer is 1.066×10^-26 kg m/s given in option (c).
(2) The wave length associated with an electron having kinetic energy 6eV is
(a) 9A.U. (b) 5A.U. (c) 2.5A.U. (d) 1.5A.U. (e) 0.5A.U.
In the case of electrons, de Broglie wave length, λ = √ [ 150/V] A.U. where V is the accelerating voltage for the electron (= 6 volt since the energy is 6 eV).
Therefore, λ = √25 = 5 A.U.
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