Thank you for reporting, we will resolve it shortly
Q.
An $\alpha$-particle moves in a circular path of radius $0.83 \,cm$ in the presence of a magnetic field of $0.25\, wb / m ^{2}$. The de-Broglie wavelength associated with the particle will be :-
Solution:
Using $\lambda=\frac{ h }{ mv }$
where $mv = qBR$
$\Rightarrow \lambda=\frac{ h }{ qBR }$
So
$\lambda=\frac{6.6 \times 10^{-34}}{\left(2 \times 1.6 \times 10^{-19}\right) \times(0.25) \times\left(0.83 \times 10^{-2}\right)}$
$=0.01 \,\mathring{A}$