InterviewSolution
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InterviewSolution
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Electrons with de - Brogli wavelengtyh `lambda` fall on the target in an X-ray tube.The cut off wavelength of the emitted Xrays is (a) `lambda_0 = (2mclambda^2)/(h)` (b)`lambda_0 = (2h)/(mc)` (c ) `lambda_0 (2m^2 c^2 lambda^3)/(h^2)` (d)`lambda_0 = lambda` |
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Answer» Correct Answer - A::B::C::D Momentum of bombarding electrons,` `p = (h)/(lambda)` `:. Kinetic energy of bombarding electrons, `K = (p^2)/(2m) = (h^2)/(2mlambda^2)` This is also maximum energy of X-ray photons. Therefore, `(hc)/(lambda_0) = (h^2)/(2mlambda^2) or `lambda_0 = (2mlambda^2 c)/(h)` :. Correct option is (a). |
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