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A proton and an alpha - particle are accelerated through same potential difference. Then, the ratio of de-Broglie wavelength of proton and alpha-particle isA. `sqrt(2):1`B. `sqrt(4):1`C. `sqrt(6):1`D. `sqrt(8):1` |
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Answer» Correct Answer - D (d) : As `lamda=(h)/(sqrt(2mqV))` `:.lamdaprop(1)/(sqrt(mq)):.(lamda_(p))/(lamda_(alpha))=(sqrt(m_(alpha)q_(alpha)))/(sqrt(m_(p)q_(p)))` `(sqrt(4m_(p)xx2e))/(sqrt(m_(p)xxe))=sqrt(8)" "( :.m_(alpha)=4m_(p),q_(alpha)=2q_(p))` |
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