Question

A uniform magnetic field with a slit system shown in fig. is to be used as a momentum filter for high energy charged particles. With a field B tesla, it is found that the filter transmits $\alpha$ -particles each of energy $5\cdot 3$ MeV. The magnetic field is increased to $2\cdot 3$ B Tesla and deuterons are passed into the filter. The energy of each deuteron transmitted by the filter is…
(A) 5MeV
(B) 7MeV
(C) 14MeV
(D) 21MeV

Answer 1

Hint : The centripetal force required by the ion to move in a circular path is provided by the perpendicular magnetic field B.
$qvB=\dfrac{m{{v}^{2}}}{r}$
m is mass of charge particle to be accelerated, B is magnetic field, q is charge on particle, v is velocity of charge particle, r is radius of circular path.
The energy of particle is given $E=\dfrac{{{p}^{2}}}{2m}$
p is the momentum of a particle.
Using the above formulas we will get the required result.

Complete step by step solution:
We have given, high energy charge particle in the magnetic field B Tesla.
Radius of charge particle is given by,
$\begin{aligned} & r=\dfrac{mv}{qB}=\dfrac{P}{qB} \\\ & \therefore r=\dfrac{\sqrt{2Em}}{Bq} \\\ \end{aligned}$
Radius in case of $\alpha$ -particles
${{r}_{\alpha }}=\dfrac{{{\sqrt{2{{E}_{\alpha }}m}}_{\alpha }}}{B{{q}_{\alpha }}}$
Radius in case of deuteron –particle is given by
${{r}_{d}}=\dfrac{{{\sqrt{2{{E}_{d}}m}}_{d}}}{2.3B{{q}_{d}}}$
Since the radius for both the particle is same
$\dfrac{{{\sqrt{2{{E}_{\alpha }}m}}_{\alpha }}}{B{{q}_{\alpha }}}=\dfrac{{{\sqrt{2{{E}_{d}}m}}_{d}}}{2\cdot 3B{{q}_{d}}}$
Squaring on both sides
$\dfrac{2{{E}_{\alpha }}{{m}_{\alpha }}}{{{q}_{\alpha }}^{2}}=\dfrac{2{{E}_{d}}{{m}_{d}}}{{{\left( 2\cdot 3 \right)}^{2}}{{q}_{d}}^{2}}$ ---- (1)
Since,
$\dfrac{{{q}_{d}}}{{{q}_{\alpha }}}=\dfrac{e}{2e}=\dfrac{1}{2}$
qd is charge on deuteron, ${{q}_{\alpha }}$ is charge on $\alpha$ -particle
$\dfrac{{{m}_{\alpha }}}{{{m}_{d}}}=\dfrac{4}{2}=2$
${{m}_{\alpha }}$ is mass of $\alpha$ -particle, ${{m}_{d}}$ is mass of deuteron
${{E}_{\alpha }}$ (energy of $\alpha$ -particle = 5.3 MeV
Put all in eq. (1)
$\begin{aligned} & {{\left( 2.3 \right)}^{2}}\times {{\left( \dfrac{{{q}_{d}}}{{{q}_{\alpha }}} \right)}^{2}}\dfrac{{{E}_{\alpha }}{{m}_{\alpha }}}{{{m}_{d}}}={{E}_{d}} \\\ & {{\left( 2.3 \right)}^{2}}\times {{\left( \dfrac{1}{2} \right)}^{2}}\times 5.3MeV(2)={{E}_{d}} \\\ \end{aligned}$
Energy of deuteron particle is given by,
$\begin{aligned} & {{E}_{d}}={{\dfrac{\left( 2.3 \right)}{4}}^{2}}\times 2\times 5.3MeV \\\ & {{E}_{d}}=14.01MeV \\\ \end{aligned}$ .

Note:
Motion of charge particle in magnetic field explained below:
Then the centripetal force acquired by charged particle to move in circular path is provided by perpendicular magnetic field B is given by
${{F}_{e}}=\dfrac{m{{v}^{2}}}{r}$ --- (1)
${{F}_{b}}=qvB$ --- (2)
From eq (1) and (2)
$\begin{aligned} & \dfrac{m{{v}^{2}}}{r}=qvB \\\ & \dfrac{mv}{r}=Bq \\\ \end{aligned}$
$r=\dfrac{mv}{qB}$ This is the radius of circular path.
The momentum of particle is given
$P=mv$
And energy of particle is,
$E=\dfrac{{{P}^{2}}}{2m}$
Or
$P=\sqrt{2mE}$ ,
Radius of charge particle is given by
$\begin{aligned} & r=\dfrac{mv}{qB}=\dfrac{P}{qB} \\\ & \therefore r=\dfrac{\sqrt{2Em}}{Bq} \\\ \end{aligned}$.