Question

A proton, a deuteron and an alpha particle are projected perpendicular to the direction of the uniform magnetic field with the same kinetic energy. The ratio of the radii of the circular paths described by them is?
A-\[\sqrt{2}:1:\sqrt{2}\]
B-\[1:\sqrt{2}:1\]
C-\[\sqrt{2}:\sqrt{2}:1\]
D-\[\sqrt{2}:1:1\]

Answer 1

Hint: All the three particles are charged particles. Proton contains positive charge +e, deuteron contains positive charge +e, and alpha particle contains charge +2e. So when they move through a region of a magnetic field they will experience a magnetic force. In order to find the ratio of the trajectory we use the equation of motion.

Complete step by step answer:
For a charged particle having charge q, moving with velocity v, and in the region containing magnetic field B, the magnetic force is given by \[\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})\]
\[\Rightarrow F=qvB\sin \delta \]
Since all of them enter perpendicularly, the angle between the velocity vector and the field vector is \[{{90}^{0}}\].
$ \Rightarrow F=qvB\sin 90 \\
 \Rightarrow F=qvB \\ $
Now given they move in a circular path, so the centripetal force must be equal to the magnetic force.
$ \Rightarrow \dfrac{m{{v}^{2}}}{r}=qvB \\
 \Rightarrow r=\dfrac{mv}{qB} \\$
Since, kinetic energy is give by,
$ E=\dfrac{m{{v}^{2}}}{2} \\
 \Rightarrow v=\dfrac{\sqrt{2E}}{\sqrt{m}} \\ $
Since the value of kinetic energy is same for everyone, thus, \[v\propto \dfrac{1}{\sqrt{m}}\]
$ \because r=\dfrac{mv}{qB} \\
 \therefore r=\dfrac{m\dfrac{\sqrt{2E}}{\sqrt{m}}}{qB} \\
 \Rightarrow r=\dfrac{\sqrt{2mE}}{qB} \\ $
Now,
Mass of proton = 1 Charge of proton = 1
Mass of deuteron = 2 Charge of deuteron = 1
Mass of alpha particle = 4 Charge of alpha particle = 2
Now for proton: \[{{r}_{P}}=\dfrac{\sqrt{2E}}{B}\]
For deuteron: \[{{r}_{d}}=\dfrac{\sqrt{4E}}{B}\]
For alpha particle: \[{{r}_{\alpha }}=\dfrac{\sqrt{8E}}{2B}\]
Now taking ratio for three of them we get,
\[{{r}_{P}}:{{r}_{d}}:{{r}_{\alpha }}=\dfrac{\sqrt{2E}}{B}:\dfrac{\sqrt{4E}}{B}:\dfrac{\sqrt{8E}}{2B}\]
\[{{r}_{P}}:{{r}_{d}}:{{r}_{\alpha }}=1:\sqrt{2}:1\]

So, the correct answer is “Option B”.

Note:
We have written here kinetic energy in terms of mass and velocity. Also we have considered the basic charge to be 1 that is the charge on the proton and considered the mass on proton as 1 that is the mass of the proton. Alpha particle is a doubly ionized helium atom. Deuterium is an isotope of hydrogen having the same atomic number but different mass number.