1.Electrons with de-Broglie wavelength λ\lambdaλ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays isa.λ0=2mcλ2h\lambda_0 = \dfrac{2mc\lambda^2}{h}λ0=h2mcλ2b.λ0=2hmc\lambda_0 = \dfrac{2h}{mc}λ0=mc2hc.λ0=2m2c2λ3h2\lambda_0 = \dfrac{2m^2c^2\lambda^3}{h^2}λ0=h22m2c2λ3d.λ0=λ\lambda_0 = \lambdaλ0=λLogin to continueOnly logged in users canattempt or see the solution.
