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A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has : (a) Greater value of de-Broglie wavelength associated with it, and (b) Less momentum ? Give reasons to justify your answer. |
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Answer» SOLUTION :(a) de-Broglie wavelength is given by `LAMBDA=(h)/(SQRT(2mqV))` As mass of proton `lt` mass of deuteron and `q_(p)=q_(d)` and v is same `implies lambda_(p) gt lambda_(d)` for same accelerating potential. (b) Momentum `= (h)/(lambda)` `:. lambda_(p) gt lambda_(d)` `:.` Momentum of proton will be less, than that of deutron. |
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