Question

An $\alpha$-particle moves in a circular path of radius $1\, cm$ in a uniform magnetic field of $0.125\, T$. The de Broglie wavelength associated with the $\alpha$-particle is

• A. $1.65 \times 10^{-12} m$
• B. $3.3 \times 10^{-12} m$
• C. $4.95 \times 10^{-12} m$
• D. $6.6 \times 10^{-12} m$

Answer 1

Answer: (A)

Given magnetic field, $B=0.125 \,T$,
Radius of the circular path, $R=1 \,cm =10^{-2} m$
We know radius of circular path in a uniform magnetic field,
$R=\frac{m v}{q B} \Rightarrow m v=q B R$
and de -Broglie wavelength, $\lambda=\frac{h}{m v}$
$\left[\because q=q_{\alpha}\right.$, For $\alpha$ - particle $]$
or $\lambda=\frac{h}{q_{\alpha} B R} $
$\lambda=\frac{6.6 \times 10^{-34}}{2 \times 1.6 \times 10^{-19} \times 0.125 \times 1 \times 10^{-2}} $
${\left[\because q_{\alpha}=+2 C =2 \times 1.6 \times 10^{-19}\right]} $
$\Rightarrow \lambda=1.65 \times 10^{-12} m$
Hence, de-Broglie wavelength associated with $\alpha$ -particle is $1.65 \times 10^{12} m$.