InterviewSolution
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InterviewSolution
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A `alpha` -parhticle moves in a circular path of radius `0.83 cm` in the presence of a magnetic field of `0.25 Wb//m^(2)`. The de-Broglie wavelength assocaiated with the particle will beA. `1Ã…`B. `0.1Ã…`C. `10Ã…`D. `0.01Ã…` |
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Answer» Correct Answer - D (d) : Radius of the circular path of a charged particle in a magnetic field is given by `R=(mv)/(Bq)ormv=RBq` Here, `R=0.83cm=0.83xx10^(-2)m` `B=0.25"Wb m"^(-2)` `q=2e=2xx1.6xx10^(-19)C` `:.mv=(0.83xx10^(-2))(0.25)(2xx1.6xx10^(-19))` de Broglie wavelength, `lamda=(h)/(mv)=(6.6xx10^(-34))/(0.83xx10^(-2)xx0.25xx2xx1.6xx10^(-19))=0.01Ã…` |
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