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For wave concerned with proton ,de-Broglie wavelength changes by 0.25 %.If its momentum changes by p_(0), initial momentum |
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Answer» `100 p_(0)` EQUATION shows that as p decreases `lambda` increases. `therefore +(0.25)/(100)lambda=(h)/(p-p_(0))` `therefore (100.25 lambda)/(100)=(h)/(p-p_(0))` .........(2) TAKING RATIO of equation (2) and (1), `(100.25)/(100)=(h)/(p-p_(0))` `therefore 100.25 p-100.25 p_(0)=100p` `therefore 0.25 p=100.25 p_(0)` `therefore p=(100.25 p_(0))/(0.25) therefore p=401 p_(0)` |
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