Question

What is the energy difference between parallel and antiparallel alignment of the zcomponent of an electron's spin magnetic dipole moment with an external magnetic field of magnitude 0.24 T, directed parallel to thezaxis?

Answer 1

Answer :

The energy difference between parallel and anti parallel alignment of the z component of an electron's spin is $4.45\times10^{-24}\ J$

Explanation :

Given that,

Magnetic field = 0.24 T

We need to calculate the energy difference between parallel and anti parallel alignment of the z component of an electron's spin

Using formula of energy difference between parallel and anti parallel alignment

$\Delta U=U_{2}-U_{1}$

$\Delta U=-\mu_{z}B\cos180^{\circ}-(-\mu_{z}B\cos0^{\circ})$

$\Delta U=2\mu_{z}B$

We know that,

The value of Bohr magneton is given by

$\mu_{z}=5.788\times10^{-5}\ eV/T$

$\mu_{z}=5.788\times10^{-5}\times1.6\times10^{-19}\ J/T$

$\mu_{z}=9.2608\times10^{-24}\ J/T$

Put the value into the formula

$\Delta U=2\times9.2608\times10^{-24}\times0.24$

$\Delta U=4.45\times10^{-24}\ J$

Hence, The energy difference between parallel and anti parallel alignment of the z component of an electron's spin is $4.45\times10^{-24}\ J$