Answer
(a) $\Delta E = 2.77 \times 10^{-7} eV$. Anti-parallel configuration has lower energy.
(b) $ f = 6.689 \times 10^7 Hz $
$\lambda = 4.48 m$
Work Step by Step
$(a)$ We use the equation $\Delta E = 2 \mu_zB$, to find the transition energy between states. where $\mu_z = 1.913 \mu_n$ t and $\mu_n = 3.1525 \times 10^{8} eV/T$
$\Delta E = 2 \mu_zB$
$\Delta E = 2 (1.913)(3.1525 \times 10^{-8} eV/T)(2.30T)$
$\Delta E = 2.77 \times 10^{-7} eV$
*The antiparallel configuration is lower energy in neutron because the $\overrightarrow μ $ and $\overrightarrow S$ are in opposite directions. This result is smaller than configuration in proton.
$(b)$ Frequency of photon
$ f = \frac{\Delta E}{h}$
$ f = \frac{(2.77 \times 10^{-7} eV)(1.6 \times 10^{-19} J/eV)}{6.626 \times10^{-34} J.s}$
$ f = 6.689 \times 10^7 Hz $ or $ 66.89 MHz$
Wavelength of photon
$\lambda = \frac{c}{f}$
$\lambda = \frac{3.0 \times 10^8 m/s }{ 6.689 \times 10^7 Hz }$
$\lambda = 4.48 m$