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

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

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$\frac{1}{2}$

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

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