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Question: An \(\alpha -\)particle moves in a circular path of radius 0.83 cm in the presence of magnetic field...

An α−\alpha -particle moves in a circular path of radius 0.83 cm in the presence of magnetic field of 0.25 Wb m-2. The de Broglie Wavelength associated with the particle will:

A

1Å

B

0.1Å

C

10Å

D

0.01Å

Answer

0.01Å

Explanation

Solution

: Radius of the circular path of a charged particle in magnetic field is given by

R=mvBqR = \frac{mv}{Bq}

Or mv=RBqmv = RBq

Here , R=0.83cm=0.83×10−2mR = 0.83cm = 0.83 \times 10^{- 2}m

B=0.25Wbm−2B = 0.25Wbm^{- 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=0.01A˚\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}} = 0.01Å