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

Question

Aicharya · Accepted Answer

We know
$R=\frac{m v}{q B}$
$\begin{aligned}
\lambda & =\frac{h}{m v} \\
\lambda & =\frac{h}{q B R} \\
& =\frac{6.6 \times 10^{-34}}{1.6 \times 10^{-19} \times 0.83 \times 10^{-2} \times 0.25} \\
& =0.01 Å
\end{aligned}$

An $\alpha$-particle moves in a circular path of radius $0.83 \mathrm{~cm}$ in the presence of a magnetic…

Solution