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Find the de Broglie wavelength of relaativistic electrons reaching the anticathode of an X-ray tube if the short wavelength limit of the continous X-rays spectrum is equal to lambda_(sh)= 10.0p m? |
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Answer» SOLUTION :For relativistic elelctrons, the formula for the short wavelength LIMIT is `X`-rays will be `(2pi ħc)/(lambda_(sh))=m_(0)c^(2)((1)/(sqrt(1-beta^(2)))-1)=csqrt(P^(2)+m^(2)c^(2))-mc^(2)` or `((2pi ħ)/(lambda_(sh))+mc)^(2)=P^(2)+m^(2)c^(2)` or `((2piħ)/(lamda_(sh)))((2piħ)/(lambda_(sh))+2MC)=P^(2)` or `p=(2piħ)/(lambda_(sh))sqrt(1+(mc lambda_(sh))/(piħ))` HENCE `lambda_(sh)=lambda_(sh)//sqrt(1+(mc lambda_(s h))/(piħ))-3.29p m` |
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