Correct Answer - Option 4 :
\(1 : √2 : 1\)
CONCEPT:
- A charged particle experiences a force when it moves in a magnetic field.
⇒ F = qvBsinθ -----(1)
Where, F = force due to magnetic field, q = magnitude of charge, v = velocity of charge, B = magnetic field, and θ = angle between v and B
- When the velocity of a charged particle is perpendicular to a magnetic field, it describes a circle and the radius of the circle is given by:
\(⇒ r=\frac{mv}{qB}\)
- Kinetic Energy: The energy possessed by a particle by the virtue of its motion is called kinetic energy. It is given by,
\(⇒ KE=\frac{1}{2}m× v^{2}\)
Where KE = kinetic energy, m = mass and v = velocity
CALCULATION:
Given:
| Parameter |
Proton |
Deuteron |
α-particle |
| Mass |
m |
2m |
4m |
| Charge |
q |
q |
2q |
| Kinetic energy |
k |
k |
k |
- The radius of the circle is given by:
\(⇒ r=\frac{mv}{qB}\) -----(1)
- The kinetic energy is given by,
\(⇒ KE=\frac{1}{2}m× v^{2}\) -----(2)
By equation 1 and equation 2,
\(\Rightarrow r=\sqrt{\frac{2km}{qB}}\) -----(3)
By equation 3 the radius of the proton is given as,
\(\Rightarrow r_{p}={\frac{\sqrt{2km}}{qB}}\) -----(4)
By equation 3 the radius of the deuteron is given as,
\(\Rightarrow r_{d}={\frac{\sqrt{2k\times2m}}{qB}}\)
\(\Rightarrow r_{d}=\sqrt{2}\times{\frac{\sqrt{2km}}{qB}}\) -----(5)
By equation 3 the radius of the α-particle is given as,
\(\Rightarrow r_{\alpha}={\frac{\sqrt{2k\times4m}}{2qB}}\)
\(\Rightarrow r_{\alpha}={\frac{\sqrt{2km}}{qB}}\) -----(6)
By equation 4, equation 5, and equation 6,
\(\Rightarrow r_{p}:r_{d}:r_{\alpha}\\\Rightarrow1:\sqrt{2}:1\)
Hence, option 4 is correct.
