The ratio of the de-Broglie wavelength of an | vedclass

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(C) When a charged particle (charge $q$,mass $m$) enters a magnetic field $B$ perpendicularly,the radius $r$ of its circular path is given by:$r = \fr

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$\frac{1}{2}$

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(C) When a charged particle (charge $q$,mass $m$) enters a magnetic field $B$ perpendicularly,the radius $r$ of its circular path is given by:$r = \frac{mv}{qB} \Rightarrow mv = qBr$The de-Broglie wavelength $\lambda$ is given by:$\lambda = \frac{h}{mv}$Substituting $mv = qBr$ into the wavelength formula:$\lambda = \frac{h}{qBr}$Given that the radii $r$ and the magnetic field $B$ are the same for both particles:$\frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{q_{p}}{q_{\alpha}}$For an $\alpha$-particle,$q_{\alpha} = 2e$,and for a proton,$q_{p} = e$:$\frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{e}{2e} = \frac{1}{2}$

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