Question #88618
a beam of electrons enters in uniform magnetic field of flux density 1.5 Tesla the energy difference between the electrons whose spin are parallel and antiparallel to the field is
1)1.74×10^-4 eV
2)0.87×10^-4 ev
3)3.48×10^-4 ev
4)2.5×10^-4 ev
1
Expert's answer
2019-04-30T09:47:23-0400

The potential energy of a particle possessing a magnetic (dipole) moment and interacting with a uniform magnetic field can be obtained as follows:


U=μB,U = - \vec{\mu} \cdot \vec{B},


where \mu is the magnetic moment of a particle which is parallel to spin. Hence, the difference between the potential energies of particles whose spin is parallel and antiparallel to the field is


ΔU=2μB\Delta U = 2 \mu B

 In case of an electron


μμB5.79105eVT,\mu \approx \mu_B \approx 5.79 \cdot 10^{-5} \, \frac{eV}{T},

where \mu_B is the Bohr magneton.

Substituting the numerical values, we obtain:


ΔU=25.791051.51.74104eV\Delta U = 2 \cdot 5.79 \cdot 10^{-5} \cdot 1.5 \approx 1.74 \cdot 10^{-4} \, eV



Answer: 1) 1.74×10-4 eV.




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS