Question

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