The ratio of de-Broglie wavelength of a $\;\alpha\;$ particle to that of proton being subjected to same magnetic field so that the radii of their path are equal to each other assuming the field induction vector $\;\overrightarrow{B}\;$ is perpendicular to velocity vectors of $\;\alpha-\;$ particle and proton , is

$(a)\;1\qquad(b)\;\large\frac{1}{4}\qquad(c)\;\large\frac{1}{2}\qquad(d)\;2$

Answer : (c) $\;\large\frac{1}{2}$
Explanation :
When a charged particle of charge q , mass m enters perpendicularly to magnetic induction B of a magnetic field , it will experience a magnetic force F=$q\;(\overrightarrow{V}\times\overrightarrow{B})$
$=qVB sin 90=qVB$
That provide a centripetal acceleration $\;\large\frac{V^{2}}{r}$
$rVB=\large\frac{mV^2}{r} \quad \; mV=qBr$
The de-Broglie wavelength
$\lambda=\large\frac{h}{mV}=\large\frac{h}{qBr}$
$\large\frac{\lambda_{\alpha-particle}}{\lambda_{photon}}=\large\frac{q_{P} r_{P}}{q_{\alpha}r_{\alpha}}$
Since $\;\large\frac{r_{\alpha}}{r_{P}}=I\;$ and $\;\large\frac{q_{\alpha}}{q_{P}}=2$
$\large\frac{\lambda_{\alpha}}{\lambda_{P}}=\large\frac{1}{2}\;.$
answered Feb 25, 2014 by 1 flag