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Question

Answers

$A)\text{ }2:3$

$B)\text{ 3}:2$

$C)\text{ 9}:4$

$D)\text{ 4}:9$

Answer
Verified

$R=\dfrac{mv}{qB}$

We will use the formula for the radius of the circular path of a charged particle moving in a magnetic field, for the two particles and get the required ratio.

Let the velocity of the first $\alpha $ - particle be ${{v}_{1}}$ and that of the second one be ${{v}_{2}}$.

Now, according to the question,

${{v}_{1}}:{{v}_{2}}=3:2$

$\therefore \dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{3}{2}$ --(1)

Now, let the radii of the circular paths of the first and second $\alpha $ - particles be ${{R}_{1}}$ and ${{R}_{2}}$ respectively.

We have to find ${{R}_{1}}:{{R}_{2}}$.

Now an $\alpha $ - particle is nothing but the nucleus of a helium atom. Let the charges on the first and second $\alpha $- particles be $q$ for both.

Let the mass of both the $\alpha $- particles be $m$.

Now, the radius $R$ of the circular path of a charge $q$ of mass $m$ moving in a magnetic field of magnitude $B$ with a speed $v$ is given by

$R=\dfrac{mv}{qB}$ --(2)

Now, using (2), we get,

${{R}_{1}}=\dfrac{m{{v}_{1}}}{qB}$ --(3)

${{R}_{2}}=\dfrac{m{{v}_{2}}}{qB}$ --(4)

Dividing (3) by (4), we get,

$\therefore \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{\dfrac{m{{v}_{1}}}{qB}}{\dfrac{m{{v}_{2}}}{qB}}=\dfrac{{{v}_{1}}}{{{v}_{2}}}$

Using (1), we get,

$\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{3}{2}=3:2$

$\therefore {{R}_{1}}:{{R}_{2}}=3:2$

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