An $α$-particle moves in a circular paths of radius 0.83 cm in the presence of a magnetic field $0.25 Wb/m^2$. The de-Broglie wavelength associated with the particle would be approximately:

0.01 Å

The correct answer is Option (4) → 0.01 Å

The radius is given by,

$R=\frac{mv}{qB}=\frac{P}{qB}$

$P=R×qB$

$=0.664×10^{-21}$

Now,

$λ=\frac{h}{p}=\frac{6.6×10^{-34}}{0.664×10^{-21}}=0.01 Å$