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A proton and an `alpha`-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths `lambda_(p) and lambda_(a)` related to each other?A. `sqrt2`B. `sqrt3`C. `sqrt8`D. `sqrt10` |
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Answer» Correct Answer - C As,`lambda=h/sqrt(2mqv),thereforelambdaprop1/sqrt(mq)` `lambda_(p)/lambda_(alpha)=sqrt(m_(alpha)q_(alpha))/sqrt(m_(p)q_(q))=sqrt(4m_(p)xx2e)/sqrt(m_(p)xxe)=sqrt8` `therefore lambda_(p)=sqrt8lambda_(alpha)` i.e., wavelength of proton is `sqrt8` times wavelength of `alpha`-particle. |
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