Electrons of mass m with de-Broglie wavelength $\lambda $ fall on the target in an X-ray tube. The cutoff wavelength ($\lambda $0) of the emitted X-ray is

Question

Electrons of mass m with de-Broglie wavelength $\lambda $ fall on the target in an X-ray tube. The cutoff wavelength ($\lambda $₀) of the emitted X-ray is

$\lambda $₀ = ${{2mc{\lambda ^2}} \over h}$

${\lambda _0} = {{2h} \over {mc}}$

${\lambda _0} = {{2{m^2}{c^2}{\lambda ^3}} \over {{h^2}}}$

${\lambda _0} = \lambda $

Solution

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