Answer
$5.79\times 10^{-5}\;eV$
Work Step by Step
The magnetic moment in Ag atoms arises due to the spin of its only one valence electron in its outer shell.
The magnetic moment of a spin-up $(+\frac{1}{2} )$ electron is $\mu_s=+1\;\mu_B$
and of a spin-down $(-\frac{1}{2} )$ electron is $\mu_s=-1\;\mu_B$
Depending upon the spin type, the Ag beam splits into two beam under external magnetic field $(\vec B)$
Thus the magnetic energy of the beam consisting of spin-up $(+\frac{1}{2})$ Ag atoms is
$U_1=\mu_BB$
and the magnetic energy of the beam consisting of spin-down $(-\frac{1}{2})$ Ag atoms is
$U_2=-\mu_BB$
Therefore, the energy difference between the magnetic moment orientations of the silver atoms in the two subbeams is
$\Delta U=U_1-U_2$
$\implies \Delta U=(\mu_BB)-(-\mu_BB)$
$\implies \Delta U=2\mu_BB$
$\implies \Delta U=2\times9.27\times 10^{-24}\times0.5\;J$
$\implies \Delta U=9.27\times 10^{-24}\;J$
$\implies \Delta U=5.79\times 10^{-5}\;eV$
