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The ratio of de·Brogfie wavelength of an particle to that of a proton being subjected to the same magnetic field so that the radii of their paths are equal to each other. assuming that the field induction vector B is perpendicular to the velocity vectors of the particle and the proton. is
detailed solution
Correct option is B
when a charged particle moves perpendicular to a magnetic field force qvB =mv2r and r= mvqB =pqB so p= qBr and de Broglie wavelength λ=hqBr here B and r are same so λ∞1 q λαλP= qPqα=12Talk to our academic expert!
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An electron is orbiting in 4th energy state in a hydrogen atom. The de-Broglie wavelength associated with this electron is nearly equal to
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