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A proton and deutron 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 :de Broglie WAVELENGTH, `lambda=h/sqrt(2mqV)` Here V same for proton and DEUTRON. As mass of proton < mass of deutron and `q_p = q_d`.Therefore, `lambda_p > lambda_d` for same accelerating potential. |
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