Home
Class 12
PHYSICS
A proton and an alpha-particle, accelera...

A proton and an `alpha`-particle, accelerated through same potential difference, enter a region of uniform magnetic field with their velocities perpendicular to the field. Compare the radii of circular paths followed by them.

Text Solution

Verified by Experts

Let mass of proton = m, charge of proton = e
Now, Mass of `alpha`-particle = 4m
Chanrge of `alpha`-particle = 2e
When a charge q is accelerated by V volts, it acquires a kinetic energy `E_(k) = qV`
`:.` Momentum is given by `mv = sqrt(2m E_(k)) = sqrt(2mqV)`
Radius `r = (mv)/(qB) or r = (sqrt(2mqV))/(qB) = sqrt((2mV)/(qB^(2))`
Thus `(r_(p))/(r_(alpha)) = sqrrt((2mV)/(eB^(2))) xx sqrt((2eB^(2))/(2(4m) V)) = (1)/(sqrt2)`
Promotional Banner

Topper's Solved these Questions

  • MOVING CHARGES AND MAGNETISM

    AAKASH INSTITUTE ENGLISH|Exercise Illustration|14 Videos

  • MOVING CHARGES AND MAGNETISM

    AAKASH INSTITUTE ENGLISH|Exercise Try Yourself|27 Videos

  • MOVING CHARGE AND MAGNESIUM

    AAKASH INSTITUTE ENGLISH|Exercise SECTION D|16 Videos

  • NUCLEI

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT (SECTION-D)|10 Videos

Similar Questions

Explore conceptually related problems

A proton and an alpha particle accellerated throught the same potential differenc e enter a region of uniform magnetic field normally if the radius of the proton orbit is 10 cm ten radius of alpha particle is

A proton, a deutron and an alpha particle, after being accelerated through the same potential difference enter a region of uniform magnetic field vecB , in a direction perpendicular to vecB . Find the ratio of their kinetic energies. If the radius of proton's circular path is 7cm , what will be the radii of the paths of deutron and alpha particle?

A proton, a deutron and an alpha - particle accelerated through the same potential difference enter a region of uniform magnetic field, moving at right angles to it. What is the ratio of their kinetic energy?

A charged particle of mass m and charge q is accelerated through a potential difference of V volts. It enters a region of uniform magnetic field which is directed perpendicular to the direction of motion of the particle. Find the radius of circular path moved by the particle in magnetic field.

A proton and alpha - particle are accelerated with same potential difference and they enter in the region of constant magnetic field B perpendicular to the velocity of particles. Find the ratio of radius of curvature of alpha - particle .

A proton and an alpha - particle are accelerated through same potential difference. Then, the ratio of de-Broglie wavelength of proton and alpha-particle is

A proton and an alpha - particle are accelerated through same potential difference. Then, the ratio of de-Broglie wavelength of proton and alpha-particle is

A proton, a deuteron and an alpha particle, are accelerated through the same potential difference and then subjected to a unifrom magnetic field vecB , perpendicular to the direction of their motions. Compare (i) their kinetic energies, and (ii) if the radius of the ciruclar path described by proton is 5 cm, determine the radii of the paths described by deuteron and alpha partcile.

An electron of mass m and charge e is accelerated by a potential difference V. It then enters a uniform magnetic field B applied perpendicular to its path. The radius of the circular path of the electron is

A proton and an alpha - Particle ( with their masses in the ratio of 1:4 and charges in the ratio of 1:2 ) are accelerated form rest through a potential differeence V, if a uniform magnetic field (B) is set up perpendicular to their velocities , the ratio of the radii r_(p) : r_(alpha) of the circular paths described by them will be :