A proton and an $lpha$-particle, having kinetic energies

Question:

A proton and an $\alpha$-particle, having kinetic energies $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\alpha}$,

respectively, enter into a magnetic field at right angles. The ratio of the

radii of trajectory of proton to that of $\alpha$-particle is $2: 1$. The ratio of

$\mathrm{K}_{\mathrm{p}}: \mathrm{K}_{\alpha}$ is :

  1. (1) $1: 8$

  2. (2) $8: 1$

  3. (3) $1: 4$

  4. (4) $4: 1$


Correct Option: , 4

Solution:

$(4)$

$\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\mathrm{p}}{\mathrm{qB}} \quad \frac{\mathrm{m}_{\alpha}}{\mathrm{m}_{\mathrm{p}}}=4$

$\frac{\mathrm{r}_{\mathrm{p}}}{\mathrm{r}_{\alpha}}=\frac{\mathrm{p}_{\mathrm{p}}}{\mathrm{q}_{\mathrm{p}}} \frac{\mathrm{q}_{\alpha}}{\mathrm{p}_{\alpha}}=\frac{2}{1}$

$\frac{\mathrm{p}_{\mathrm{p}}}{\mathrm{p}_{\alpha}}=\frac{2 \mathrm{q}_{\mathrm{p}}}{\mathrm{q}_{\alpha}}=2\left(\frac{1}{2}\right)$

$\frac{p_{p}}{p_{a}}=1$

$\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{K}_{\alpha}}=\frac{\mathrm{p}_{\mathrm{p}}^{2}}{\mathrm{p}_{\alpha}^{\mathrm{p}}} \frac{\mathrm{m}_{\alpha}}{\mathrm{m}_{\mathrm{p}}}=(1)(4)$

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