For a specific material, the energies and wavelengths of its...

The energy of characteristic X-rays is given by the difference in energy levels of the atom: $E = E_{n_2} - E_{n_1}$.

The energy levels of an atom are approximately given by $E_n \propto -\dfrac{1}{n^2}$.

For the $K$ series, the transitions are to the $K$ shell ($n=1$):
$E(K_\alpha) = E_2 - E_1$
$E(K_\beta) = E_3 - E_1$
$E(K_\gamma) = E_4 - E_1$

Since $E_4 > E_3 > E_2$, we have $E(K_\gamma) > E(K_\beta) > E(K_\alpha)$.

Using the relation $\lambda = \dfrac{hc}{E}$, the wavelengths follow the reverse order:
$\lambda(K_\alpha) > \lambda(K_\beta) > \lambda(K_\gamma)$

For the $\alpha$ transitions of different series, the transitions are between adjacent shells:
$E(K_\alpha) = E_2 - E_1$
$E(L_\alpha) = E_3 - E_2$
$E(M_\alpha) = E_4 - E_3$

Since the energy gap between adjacent levels decreases as $n$ increases, we have:
$E(K_\alpha) > E(L_\alpha) > E(M_\alpha)$

Correspondingly, the wavelengths follow the reverse order:
$\lambda(M_\alpha) > \lambda(L_\alpha) > \lambda(K_\alpha)$