Answer: Option B
:
B
In the hint given we asked you to think about two questions:
1) How minimum wavelength is affected by accelerating potential and atomic number?
2) How wavelength of characteristic $K_{\alpha }$ x-ray is affected by accelerating potential?
Let's see whether answering these two leads us to our solution or not ?
Let's answer the 1st question first.
Minimum wavelength is given by ${\lambda }_0$ = $\frac{hc}{e{V}_0}$.
We see that it only depends on the accelerating potential but not the atomic number.
Also, a lesser ${\lambda }_0$ implies to a higher accelerating potential, hence
$V_A$ > $V_B$.
Now let's answer the 2nd question.
$K_{\alpha }$ refers to the transition of electron from L shell to K . In the process energy is released. If the Z is greater, then the force of attraction for this electron will be higher hence we can conclude that more energy will be released.
It’s also evident from the Moseley's law using which we see that the wavelength of $K_{\alpha }$ x - ray depends on atomic number but is not related to accelerating potential.
The relation is given by:
$\sqrt{v}=a(Z-b)OR\sqrt{\frac{c}{\lambda }}=a(Z-b)$
We find that a lesser $K_{\alpha }$ Characteristic x-ray wavelength implies a higher atomic number.
From the graph;
$Z_A$ < $Z_B$.