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(D) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB}$,where $m$ is the mass,$v$ is the velocity,$q$ is the c

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$2.1 \ \mathring{A}$

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(D) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB}$,where $m$ is the mass,$v$ is the velocity,$q$ is the charge,and $B$ is the magnetic field strength.From this,the momentum $p = mv = qBr$.The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{qBr}$.For an alpha particle,the charge $q = 2e = 2 	imes 1.6 	imes 10^{-19} \ C = 3.2 	imes 10^{-19} \ C$.Given: $r = 0.5 \ mm = 0.5 	imes 10^{-3} \ m$,$B = 2 	imes 10^{-2} \ T$,and $h = 6.63 	imes 10^{-34} \ J \ s$.Substituting these values:$\lambda = \frac{6.63 	imes 10^{-34}}{(3.2 	imes 10^{-19}) 	imes (2 	imes 10^{-2}) 	imes (0.5 	imes 10^{-3})}$$\lambda = \frac{6.63 	imes 10^{-34}}{3.2 	imes 10^{-24}} \approx 2.07 	imes 10^{-10} \ m$.Since $1 \ \mathring{A} = 10^{-10} \ m$,we have $\lambda \approx 2.1 \ \mathring{A}$.

An alpha particle moves along a circular path of radius $0.5 \ mm$ in a magnetic field of $2 \times 10^{-2} \ T$. The de Broglie wavelength associated with the alpha particle is nearly (Planck's constant $= 6.63 \times 10^{-34} \ J \ s$)

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