A deuteron and an alpha particle having equal kinetic energy
Question:

A deuteron and an alpha particle having equal kinetic energy enter perpendicular into a magnetic field. Let $r_{d}$ and $r_{\alpha}$ be their respective radii of

circular path. The value of $\frac{r_{d}}{r_{\alpha}}$ is equal to :

  1. $\frac{1}{\sqrt{2}}$

  2. $\sqrt{2}$

  3. 1

  4. 2


Correct Option: , 2

Solution:

$r=\frac{m v}{q B}=\frac{\sqrt{2 m k}}{q B}$

$\frac{\mathrm{r}_{\mathrm{d}}}{\mathrm{r}_{\alpha}}=\sqrt{\frac{\mathrm{m}_{\mathrm{d}}}{\mathrm{m}_{\alpha}} \frac{\mathrm{q}_{\mathrm{\alpha}}}{\mathrm{q}_{\mathrm{d}}}}=\sqrt{\frac{2}{4}}\left(\frac{2}{1}\right)=\sqrt{2}$

Hence option (2).

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