Question 161: Magnetism and Matter

Question

A deuteron and an alpha particle having equal kinetic energy enter perpendicularly into a magnetic field. Let rd and r$$\alpha$$ be their respective radii of circular path. The value of $${{{r_d}} \over {{r_\alpha }}}$$ is equal to :

Accepted Answer

Given, kinetic energy of $$\alpha$$-particle (K
$$\alpha$$
) = kinetic energy of deuteron (K
d
)
Since, kinetic energy, K = $${1 \over 2}$$ mv
2
$$\Rightarrow$$ mv
2
= 2K $$\Rightarrow$$ v
2
= $${{2K} \over m}$$
$$\Rightarrow$$ v = $$\sqrt {{{2K} \over m}} $$ .... (i)
We know that,
r = $${{mv} \over {Bq}}$$ .... (ii)
where, r = radius of curvature of path of a charged particle, m = mass of the charged particle, q = charge of the particle, v = velocity of charged particle and B = magnetic field.
From Eqs. (i) and (ii), we get
$$r = {{m\sqrt {{{2K} \over m}} } \over {Bq}}$$
$$ \Rightarrow r = {{\sqrt {2Km} } \over {Bq}}$$ .... (iii)
Since, m, K and B are same for both deuteron and $$\alpha$$-particle.
From Eq. (iii), we get
$$\gamma  \propto {{\sqrt m } \over q}$$
$$	herefore$$ $${{{r_d}} \over {{r_\alpha }}} = \sqrt {{{{m_d}} \over {{m_\alpha }}}} .{{{q_\alpha }} \over {{q_d}}} = \sqrt {{2 \over 4}} \left( {{2 \over 1}} ight)$$
[$$\because$$ m
d
= 2m
p
and m
$$\alpha$$
= 4m
p
q
$$\alpha$$
= 2e and q
d
= e.]
$${{{r_d}} \over {{r_\alpha }}} = \sqrt 2 $$

Answer

1

Answer

2

Answer

$$\sqrt 2 $$

Answer

$${1 \over {\sqrt 2 }}$$

Magnetism and Matter

A deuteron and an alpha particle having equal kinetic energy enter perpendicularly into a magnetic field. Let rd and r$\alpha be their respective radii of circular path. The value of {{{r_d}} \over {{r_\alpha }}}$ is equal to :

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