Answer
$57.51$
Work Step by Step
In the previous section of this problem, we have calculated that the ratio of the photon energies due to $K_{\alpha}$ transitions in two atoms whose atomic numbers are $Z$ and $Z^{\prime}$ is $\frac{(Z-1)^2}{(Z^{\prime}-1)^2}$
In the above ration, substituting the atomic numbers $Z=92$ for uranium and $Z^{\prime}=13$ for aluminum, we obtain
$\frac{(Z-1)^2}{(Z^{\prime}-1)^2}=\frac{(92-1)^2}{(13-1)^2}\approx57.51$
Therefore, the ratio for uranium and aluminum is $57.51$
