# An $\alpha$ particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m2. The de Broglie wavelength associated with the particle will be: Option 1) $1\: \: A$ Option 2) $0.1\: \: A$ Option 3) $10\: \: A$ Option 4) $0.01\: \: A$

As we learnt in

De - Broglie wavelength -

$\lambda = \frac{h}{p}= \frac{h}{mv}= \frac{h}{\sqrt{2mE}}$

- wherein

$h= plank's\: constant$

$m= mass \: of\: particle$

$v= speed \: of\: the \: particle$

$E= Kinetic \: energy \: of \: particle$

Radius in a magnetic field is $R=\frac{mv}{qB}\:\: \: or\:\: \: mv=qBR$

the debroglic wavelength $\lambda=\frac{h}{mv}=\frac{h}{qBR}=\frac{6.62 \times 10^{-34}}{(2\times 1.6 \times 10^{-19}\times 0.25 \times 8.3\times 10^{-3})}$

$\lambda=\frac{6.62 \times 10 ^{-34}}{6.64 \times 10^-22} \simeq 10^{-12} m=0.01\AA$

Option 1)

$1\: \: A$

Incorrect

Option 2)

$0.1\: \: A$

Incorrect

Option 3)

$10\: \: A$

Incorrect

Option 4)

$0.01\: \: A$

Correct

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