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

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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 $\;\large\frac{q_{\alpha}}{q_{P}}=2$

$\large\frac{\lambda_{\alpha}}{\lambda_{P}}=\large\frac{1}{2}\;.$

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