
Which particles will have minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field
A . ${Li^+}$
B . Electron
C . Proton
D . ${He^+}$
Answer
161.1k+ views
Hint: In this question we have to find the frequency of revolution of lithium ions, an electron, a proton and a helium ion when projected with the same velocity perpendicular to a magnetic field. There is a centripetal force operating on a charged particle that revolves in a circular path in a uniform magnetic field, and that force is equal to the magnetic force acting on the particle.
Formula used:
Magnetic force on the particle:
${F_m}$= q(v×B); here, q denotes the charge, v is the velocity of the particle, and B the magnetic field.
The centripetal force is given by,
${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$
Here m is the mass of the particle, v is the velocity of the particle, and r is the radius.
The expression for frequency:
$f=\dfrac{v}{2\pi r}$; here, v is the velocity of the particle and r is the radius.
Complete answer:
Magnetic force on the particle:
${F_m}$= q(v×B); here, q denotes the charge, v is the velocity of the particle, and B the magnetic field.
The centripetal force is given by,
${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$
Here m is the mass of the particle, v is the velocity of the particle, and r is the radius.
The centripetal force counterbalances the magnetic pull that acts on a moving particle perpendicular to its velocity. The equilibrium formula states the following:
${F_m}$(magnetic force)=${F_c}$(centripetal force) - (i)
$qvB=\dfrac{m{{v}^{2}}}{r}$
$\dfrac{v}{r}=\dfrac{qB}{m}$
We know that the formula for frequency is:
$f=\dfrac{v}{2\pi r}=\dfrac{qB}{2\pi m}$
The frequency of a particle depends on its mass, charge and the magnetic field.
The value of the magnetic field and charge is the same for all the particles. Therefore, the frequency depends on the mass of the particle inversely.
$f\alpha \dfrac{1}{m}$
${F_{minimum}}$ will occur for a particle with maximum mass.
Among the given particles, ${Li^+}$ have the maximum mass and thus will have the minimum frequency.
The correct answer is A.
Note: ${Li^+}$ ion has 3 protons and 4 neutrons and ${He^+}$ ion has 2 protons and 2 neutrons making ${Li^+}$ ion the heaviest.
When a charged particle of charge 'q' and mass 'm' is projected to a magnetic field perpendicular to the magnetic field then it revolves with a certain frequency which depends on the mass and charge of the particle and the magnetic field.
Formula used:
Magnetic force on the particle:
${F_m}$= q(v×B); here, q denotes the charge, v is the velocity of the particle, and B the magnetic field.
The centripetal force is given by,
${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$
Here m is the mass of the particle, v is the velocity of the particle, and r is the radius.
The expression for frequency:
$f=\dfrac{v}{2\pi r}$; here, v is the velocity of the particle and r is the radius.
Complete answer:
Magnetic force on the particle:
${F_m}$= q(v×B); here, q denotes the charge, v is the velocity of the particle, and B the magnetic field.
The centripetal force is given by,
${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$
Here m is the mass of the particle, v is the velocity of the particle, and r is the radius.
The centripetal force counterbalances the magnetic pull that acts on a moving particle perpendicular to its velocity. The equilibrium formula states the following:
${F_m}$(magnetic force)=${F_c}$(centripetal force) - (i)
$qvB=\dfrac{m{{v}^{2}}}{r}$
$\dfrac{v}{r}=\dfrac{qB}{m}$
We know that the formula for frequency is:
$f=\dfrac{v}{2\pi r}=\dfrac{qB}{2\pi m}$
The frequency of a particle depends on its mass, charge and the magnetic field.
The value of the magnetic field and charge is the same for all the particles. Therefore, the frequency depends on the mass of the particle inversely.
$f\alpha \dfrac{1}{m}$
${F_{minimum}}$ will occur for a particle with maximum mass.
Among the given particles, ${Li^+}$ have the maximum mass and thus will have the minimum frequency.
The correct answer is A.
Note: ${Li^+}$ ion has 3 protons and 4 neutrons and ${He^+}$ ion has 2 protons and 2 neutrons making ${Li^+}$ ion the heaviest.
When a charged particle of charge 'q' and mass 'm' is projected to a magnetic field perpendicular to the magnetic field then it revolves with a certain frequency which depends on the mass and charge of the particle and the magnetic field.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE General Topics in Chemistry Important Concepts and Tips

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

If a wire of resistance R is stretched to double of class 12 physics JEE_Main
