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(D) The radius $R$ of the circular path of a charged particle in a magnetic field is given by $R = \frac{mv}{Bq}$,which implies $mv = RBq$.Given value

Vedclass · Accepted Answer

$0.01$

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(D) The radius $R$ of the circular path of a charged particle in a magnetic field is given by $R = \frac{mv}{Bq}$,which implies $mv = RBq$.Given values:$R = 0.83\, cm = 0.83 	imes 10^{-2}\, m$$B = 0.25\, Wb/m^2$$q = 2e = 2 	imes 1.6 	imes 10^{-19}\, C$Calculating the momentum $p = mv$:$mv = (0.83 	imes 10^{-2}) 	imes (0.25) 	imes (2 	imes 1.6 	imes 10^{-19})$$mv = 0.83 	imes 10^{-2} 	imes 0.25 	imes 3.2 	imes 10^{-19}$$mv = 0.664 	imes 10^{-21}\, kg\cdot m/s$The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$,where $h = 6.63 	imes 10^{-34}\, J\cdot s$ (using $6.6$ for approximation as per the provided solution context):$\lambda = \frac{6.6 	imes 10^{-34}}{0.664 	imes 10^{-21}} \approx 9.94 	imes 10^{-13}\, m \approx 10^{-12}\, m$.Converting to $\mathring{A}$ $(1\, \mathring{A} = 10^{-10}\, m)$:$\lambda = 0.01 	imes 10^{-10}\, m = 0.01\, \mathring{A}$.

An $\alpha$-particle moves in a circular path of radius $0.83\, cm$ in the presence of a magnetic field of $0.25\, Wb/m^2$. The de Broglie wavelength associated with the particle will be .............. $\mathring{A}$.

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