Question 225: Magnetism and Matter

Question

An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of
radii r
e
, r
p
, r$$_\alpha$$ respectively in a uniform magnetic field B. The relation between r
e
, r
p
, r$$_\alpha$$ is:

Accepted Answer

When a  charged particle moves in a magnetic field then the charged particle moves in a circular path.  So,
$${{m{v^2}} \over r}$$ = Bqv
$$ \Rightarrow $$$$\,\,\,$$ r = $${{mv} \over {Bq}}$$
We know kinetic energy,  K = $${1 \over 2}$$ mv
2
$$	herefore\,\,\,$$ mv = $$\sqrt {2Km} $$
$$	herefore\,\,\,$$ r = $${{\sqrt {2Km} } \over {Bq}}$$
According to the question,
K
e
(electron) = K
p
(proton) = K
$$\alpha $$
(Alpha particle) = K = constant, and all of them are in uniform magnetic field.
$$ 	herefore $$ B = constant.
$$	herefore\,\,\,$$ r $$ \propto $$ $${{\sqrt m } \over q}$$
For proton (
1
H
1
), mass = m, and charge = e
$$	herefore\,\,\,$$ r
p
$$ \propto $$ $${{\sqrt m } \over e}$$
For alpha particle (
2
H
4
),
mass = 4m
and charge = 2e
$$	herefore\,\,\,$$ r
$$ \alpha $$
$$ \propto $$ $${{\sqrt {4m} } \over {2e}}$$ $$ \propto $$ $${{\sqrt m } \over e}$$
$$ 	herefore $$$$\,\,\,$$ r
p
= r
$$ \alpha$$
For electron,
charge = e
and mass  (m
e
) = 9.1 $$ 	imes $$ 10
$$-$$31
kg
and mass of proton = 1.67 $$ 	imes $$ 10
$$-$$27
kg
$$	herefore\,\,\,$$ mass of electron < mass of proton.
r
e
$$ \propto $$ $${{\sqrt {{m_e}} } \over e}$$ < r
p
$$	herefore\,\,\,$$ r
e
< r
p
= r
$$ \propto $$

Answer

r
e
< r$$_\alpha$$ < r
p

Answer

r
e
> r
p
= r$$_\alpha$$

Answer

r
e
< r
p
= r$$_\alpha$$

Answer

r
e
< r
p
< r$$_\alpha$$

Magnetism and Matter

An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii r e , r p , r$_\alpha respectively in a uniform magnetic field B. The relation between r e , r p , r_\alpha$ is:

Choose the correct answer: