Answer
$2070.25$
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 relation, substituting the atomic numbers $Z=92$ for uranium and $Z^{\prime}=3$ for lithium, we obtain
$\frac{(Z-1)^2}{(Z^{\prime}-1)^2}=\frac{(92-1)^2}{(3-1)^2}=2070.25$
Therefore, the ratio for uranium and lithium is $2070.25$
