From the information below, identify element X.a. The wavelength of the radio waves sent by an FM station broadcasting at $97.1\mathrm{MHz}$ is 30.0 million $\left(3.00\times 10^7\right)$ times greater than the wavelength corresponding to the energy difference between a particular excited state of the hydrogen atom and the ground state.b. Let $V$ represent the principal quantum number for the valence shell of element X. If an electron in the hydrogen atom falls from shell $V$ to the inner shell corresponding to the excited state mentioned above in part a, the wavelength of light emitted is the same as the wavelength of an electron moving at a speed of $570.\mathrm{m}/\mathrm{s}$c. The number of unpaired electrons for element $X$ in the ground state is the same as the maximum number of electrons in an atom that can have the quantum number designations $n=2,m_{\ell }=-1,$ and $m_s=-\frac{1}{2}$d. Let $A$ equal the principal quantum number for the lowest energy excited state for hydrogen. This value of $A$ also represents the angular momentum quantum number for the subshell containing the unpaired electron(s) for element X.