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Physics Question on Dual nature of matter

An α−\alpha - particle moves in a circular path of radius 0.83 cm0.83\, cm in the presence of a magnetic field of 0.25 Wb/m2.0.25\, Wb/m^2. The de Broglie wavelength associated with the particle will be (Take h=6.63?10−34J sh = 6.63 ? 10^{-34} J\, s, e=1.6?10−19Ce =1.6 ?10^{-19}C)

A

1 ?1\, ?

B

0.1 ?0.1 \, ?

C

10 ?10\, ?

D

0.01 ?0.01\, ?

Answer

0.01 ?0.01\, ?

Explanation

Solution

Radius of the circular path of a charged particle in a magnetic field is given by
R=mvBq  mv=RBqR= \frac {mv}{Bq} \,\,mv=RBq
Here, R=0.83 cm=0.83×10−2mR=0.83\, cm=0.83 \times 10 ^{-2}m
B=0.25 Wb m2B=0.25 \, Wb \, m^2
q=2e=2×1.6×10−19Cq=2e=2 \times 1.6 \times 10^{-19}C
∴mv=(0.83×10−2)(0.25)(2×1.6×10−19)\therefore mv=(0.83 \times 10^{-2})(0.25)(2 \times 1.6 \times 10^{-19})
de Broglie wavelength,
λ=hmv=6.6×10−340.83×10−2×0.25×2×1.6×10−19\lambda=\frac {h}{mv}= \frac {6.6 \times 10^{-34}}{0.83 \times 10^{-2}\times 0.25 \times 2 \times 1.6 \times 10^{-19}}
=0.01 ?=0.01 \,?