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InterviewSolution
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The ratio of de - Broglie wavelength of `alpha `- particle to that of a proton being subjected to the same magnetic field so that the radii of their path are equal to each other assuming the field induction vector `vec(B)` is perpendicular to the velocity vectors of the `alpha` - particle and the proton isA. `1`B. `1//4`C. `1//2`D. `2` |
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Answer» When a charged particle of charge `q`, mass `m` enters perpendicularly to the magnetic induction `vecB` of a magnetic field, it will experience a magnetic force `F=q(vecvxxvecB)=q vB sin 90^(@)=qvB` that provide a centripetal acceleration `(v^(2))/(r )` `rArr qvB=(mv^(2))/(r )rArrmv=qBr` `rArr` The `de`-Broglie wavelength `lambda=(h)/(mv)=(h)/(qBr)` `rArr (lambda_(alpha-"particle"))/(lambda_("proton"))=(q_(p)r_(p))/(q_(alpha)r_(alpha))` Since `(r_(alpha))/(r_(p))=1` and `(q_(alpha))/(q_(p))=2` `rArr (lambda_(alpha))/(lambda_(p))=1//2` |
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