A proton moves in a circular path of radius 6 | vedclass

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Question

(C) Given: Radius $r = 6.6 	imes 10^{-3} \ m$,Magnetic field $B = 0.625 \ T$,Charge of proton $q = 1.6 	imes 10^{-19} \ C$,Planck's constant $h = 6.

Vedclass · Accepted Answer

$0.01$

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(C) Given: Radius $r = 6.6 	imes 10^{-3} \ m$,Magnetic field $B = 0.625 \ T$,Charge of proton $q = 1.6 	imes 10^{-19} \ C$,Planck's constant $h = 6.6 	imes 10^{-34} \ J \cdot s$.For a charged particle in a magnetic field,the magnetic force provides the centripetal force: $\frac{mv^2}{r} = qvB$.Thus,momentum $p = mv = qBr$.Substituting the values: $p = (1.6 	imes 10^{-19} \ C) 	imes (0.625 \ T) 	imes (6.6 	imes 10^{-3} \ m) = 6.6 	imes 10^{-22} \ kg \cdot m/s$.The de Broglie wavelength is given by $\lambda = \frac{h}{p}$.$\lambda = \frac{6.6 	imes 10^{-34}}{6.6 	imes 10^{-22}} = 10^{-12} \ m$.Since $1 \ \mathring{A} = 10^{-10} \ m$,then $10^{-12} \ m = 0.01 \ \mathring{A}$.

$A$ proton moves in a circular path of radius $6.6 \times 10^{-3} \ m$ perpendicular to a magnetic field of $0.625 \ T$. Find the de Broglie wavelength associated with the proton in $\mathring{A}$.

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