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Electrons with de Broglie wavelength a fall on the target in an X-ray tube. Find the cutoff wavelength of the emitted X-rays. |
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Answer» Solution :If `lamda`=de Broglie wavelength, then momentum `p=(h)/(lamda)` K.E of the STRIKING electrons `K=(p^(2))/(2m)=(h^(2))/(2mlamda^(2))` This is equal to MAXIMUM energy of X - ray photons. `(hc)/(lamda_(0))=(h^(2))/(2mlamda^(2))implieslamda_(0)=(2mlamda^(2)C)/(h)` |
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