Question

An $\alpha$ particle moves in circular path of radius $0.83cm$ in the presence of a magnetic field of $0.25Wb/m^2$. Find the De Broglie wavelength associated with the particle

Answer 1

We know that,

Radius of the circular path $r=\frac{mv}{qB}=\frac{p}{qB}$

• $p$ is momentum of the charged particle.
• For $\alpha$ particle $q=3.2\times {10}^{-19}C$ .
• $B$ is given as $0.25$ all in SI units

So momentum $p=rqB$
put $r=.0083meter$ we get De Broglie wavelength

$\lambda =\frac{h}{p}=9.939\times {10}^{-13}meter$

where $h$ is Planck constant whose value is $6.6\times {10}^{-34}Joule-second$.