Question

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

• A. $0.1\stackrel{\circ }{\text{A}}$
• B. $0.01\stackrel{\circ }{\text{A}}$
• C. $10\stackrel{\circ }{\text{A}}$
• D. $1\stackrel{\circ }{\text{A}}$

Answer 1

The correct option is B $0.01\stackrel{\circ }{\text{A}}$

The magnetic force provides the necessary centripetal force acting on the charge,
$\frac{m{v}^2}{R}=qvB$

$mv=qBR$

de-Broglie wavelength is given by,
$\lambda =\frac{h}{mv}$

$\lambda =\frac{h}{qBR}$

Substituting the values, we get,

$\lambda =\frac{6.6\times {10}^{-24}}{2\times 1.6\times {10}^{-19}\times .25\times 0.83\times {10}^{-2}}$

$\therefore \lambda =0.01\stackrel{\circ }{\text{A}}$

Hence, option $\left(B\right)$ is correct.