Answer
$1.40\times 10^{10}\;Hz$
Work Step by Step
In the previous section of this problem, we have calculated that the energy difference between the magnetic moment orientations of the silver atoms in the two subbeams is given by $\Delta U=5.79\times 10^{-5}\;eV=9.27\times 10^{-24}\;J$
Let, $f$ be the frequency of the radiation that would induce a transition between
the above two states. Therefore energy of the radiation is
$E=hv$
The condition of transition between these two states is
$E=\Delta U$
or, $hv=9.27\times 10^{-24}$
or, $v=\frac{9.27\times 10^{-24}}{6.63\times 10^{-34}}\;Hz$
or, $v\approx1.40\times 10^{10}\;Hz$
Therefore, the frequency of the radiation is $1.40\times 10^{10}\;Hz$
