The provided figure illustrates the intensity–wavelength relations...

From the given graph, the cut-off wavelength (minimum wavelength) for tube A (solid curve) is smaller than that for tube B (dashed curve):

$\lambda_{min, A} < \lambda_{min, B}$

The cut-off wavelength is given by \lambda_\min = \dfrac{hc}{eV}. Since \lambda_\min is inversely proportional to the accelerating voltage $V$, we get:

$V_A > V_B$

Also, from the graph, the wavelength of the $K_\alpha$ peak for tube A is greater than that for tube B:

$\lambda_{K_\alpha, A} > \lambda_{K_\alpha, B}$

According to Moseley's law, the wavelength of characteristic X-rays is related to the atomic number $Z$ by $\dfrac{1}{\lambda} \propto (Z - b)^2$. A larger wavelength corresponds to a smaller atomic number. Thus:

$Z_A < Z_B$

Therefore, $V_A > V_B$ and $Z_A < Z_B$.