Alpha particles,each having a charge of +2e and a mass of 6.64 × 10^-27 kg,are accelerated in a uniform

Alpha particles,each having a charge of +2e and a mass of 6.64 × 10<sup>-27</sup> kg,are accelerated in a uniform Magnetic field to a final orbit radius of The field is perpendicular to the velocity of the particles.What is the kinetic energy of an alpha particle in the final orbit? (1 eV = 1.60 × 10<sup>-19</sup> J, A)3.0 MeV B)2.6 MeV C)3.4 MeV D)3.9 MeV E)4.3 MeV

Magnetic field to a final orbit radius of Alpha particles,each having a charge of +2e and a mass of 6.64 × 10-27 kg,are accelerated in a uniform Magnetic field to a final orbit radius of The field is perpendicular to the velocity of the particles.What is the kinetic energy of an alpha particle in the final orbit? (1 eV = 1.60 × 10-19 J, A)3.0 MeV B)2.6 MeV C)3.4 MeV D)3.9 MeV E)4.3 MeV

The field is perpendicular to the velocity of the particles.What is the kinetic energy of an alpha particle in the final orbit? (1 eV = 1.60 × 10^-19 J, Alpha particles,each having a charge of +2e and a mass of 6.64 × 10-27 kg,are accelerated in a uniform Magnetic field to a final orbit radius of The field is perpendicular to the velocity of the particles.What is the kinetic energy of an alpha particle in the final orbit? (1 eV = 1.60 × 10-19 J, A)3.0 MeV B)2.6 MeV C)3.4 MeV D)3.9 MeV E)4.3 MeV

A)3.0 MeV
B)2.6 MeV
C)3.4 MeV
D)3.9 MeV
E)4.3 MeV

A Flat Circular Loop of Wire of Radius 0

Multiple Choice

Correct Answer:VerifiedUnlock this answer nowGet Access to more Verified Answers free of chargeAccess For Free

Correct Answer:
Verified
Unlock this answer now
Get Access to more Verified Answers free of charge