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Aipha particle m= 6.7×10^-27,q= 2e are accelerated rest through potential difference of 6.7kv they are enter a magnetic field B=0.2T perpendicular to their direction of motion the radius of the path described by them is fieB=?
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Aipha particle m= 6.7×10^-27,q= 2e are accelerated rest through potent...
**Acceleration of Alpha Particles**
The given problem involves an alpha particle with a mass of m = 6.7×10^-27 kg and a charge of q = 2e, where e is the elementary charge. The alpha particle is accelerated from rest through a potential difference of 6.7 kV.

To determine the acceleration of the alpha particle, we can use the equation:

acceleration = (charge * potential difference) / mass

Substituting the values given, we have:

acceleration = (2e * 6.7 kV) / 6.7×10^-27 kg

Using the elementary charge value of e = 1.6×10^-19 C, we can convert the potential difference to volts:

acceleration = (2 * 1.6×10^-19 C * 6.7×10^3 V) / 6.7×10^-27 kg

Simplifying the expression, we get:

acceleration = 6.4×10^7 m/s²

**Motion of the Alpha Particles**
The alpha particles are then subjected to a magnetic field B = 0.2 T that is perpendicular to their direction of motion. When a charged particle moves through a magnetic field, it experiences a force known as the Lorentz force.

The Lorentz force on a charged particle moving through a magnetic field is given by the equation:

force = charge * velocity * magnetic field

Since the alpha particles are moving perpendicular to the magnetic field, the force acts as a centripetal force and is given by:

force = mass * acceleration = (m * v²) / r

where v is the velocity of the alpha particle and r is the radius of the path described by the alpha particle.

**Radius of the Path**
Equating the two expressions for the force, we have:

(m * v²) / r = (charge * velocity * magnetic field)

Simplifying the equation, we find:

r = (m * v) / (charge * magnetic field)

Substituting the given values, we get:

r = (6.7×10^-27 kg * v) / (2 * 1.6×10^-19 C * 0.2 T)

Simplifying further, we have:

r = (v * 10^8 m)

Therefore, the radius of the path described by the alpha particles is 10^8 times the velocity of the particles.

**Conclusion**
In conclusion, the radius of the path described by the alpha particles is given by r = v * 10^8 m. The acceleration of the alpha particles can be calculated using the equation acceleration = (charge * potential difference) / mass. The motion of the alpha particles in a magnetic field is governed by the Lorentz force, and the radius of the path is determined by equating the centripetal force to the Lorentz force.
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Aipha particle m= 6.7×10^-27,q= 2e are accelerated rest through potential difference of 6.7kv they are enter a magnetic field B=0.2T perpendicular to their direction of motion the radius of the path described by them is fieB=?
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