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(A) The radius of a circular path for a charged particle in a uniform magnetic field is given by $R = \frac{mv}{qB}$.From this,the momentum $p = mv =

Vedclass · Accepted Answer

$1.65 	imes 10^{-12} \ m$

Vedclass · Answer

(A) The radius of a circular path for a charged particle in a uniform magnetic field is given by $R = \frac{mv}{qB}$.From this,the momentum $p = mv = qRB$.The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{qRB}$.For an $\alpha$-particle,the charge $q = +2e = 2 	imes 1.6 	imes 10^{-19} \ C$.Given $R = 1 \ cm = 10^{-2} \ m$ and $B = 0.125 \ T$.Substituting the values:$\lambda = \frac{6.6 	imes 10^{-34}}{2 	imes 1.6 	imes 10^{-19} 	imes 0.125 	imes 10^{-2}}$$\lambda = \frac{6.6 	imes 10^{-34}}{0.4 	imes 10^{-21}}$$\lambda = 1.65 	imes 10^{-12} \ m$.

An $\alpha$-particle moves in a circular path of radius $1 \ cm$ in a uniform magnetic field of $0.125 \ T$. The de Broglie wavelength associated with the $\alpha$-particle is

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