Aluminium ( Z = 13 ) and zinc ( Z = 30 ) emit K_α X-rays with...

According to Moseley's law, the frequency $\nu$ of the emitted X-ray is given by:
$\sqrt{\nu} = a(Z - b)$

Since $\nu = \dfrac{c}{\lambda}$, we can write this in terms of wavelength $\lambda$ as:
$\dfrac{1}{\sqrt{\lambda}} = k(Z - b)$
where $k = \dfrac{a}{\sqrt{c}}$. This shows that $\dfrac{1}{\sqrt{\lambda}}$ is a linear function of the atomic number $Z$.

Given the data for Aluminium ($Z = 13$) and Zinc ($Z = 30$):
$\dfrac{1}{\sqrt{887}} = k(13 - b)$ ... (i)
$\dfrac{1}{\sqrt{146}} = k(30 - b)$ ... (ii)

For Iron ($Z = 26$), let the wavelength be $\lambda_{Fe}$:
$\dfrac{1}{\sqrt{\lambda_{Fe}}} = k(26 - b)$ ... (iii)

Subtracting equation (i) from (ii) gives:
$k(30 - 13) = \dfrac{1}{\sqrt{146}} - \dfrac{1}{\sqrt{887}}$
$17k \approx 0.08276 - 0.03358 = 0.04918$
$k \approx 0.002893 \text{ pm}^{-1/2}$

Now, we can find the value for Iron by subtracting $4k$ from the Zinc equation:
$k(26 - b) = k(30 - b) - 4k$
$\dfrac{1}{\sqrt{\lambda_{Fe}}} = \dfrac{1}{\sqrt{146}} - 4k$
$\dfrac{1}{\sqrt{\lambda_{Fe}}} \approx 0.08276 - 4(0.002893)$
$\dfrac{1}{\sqrt{\lambda_{Fe}}} \approx 0.08276 - 0.01157 = 0.07119 \text{ pm}^{-1/2}$

Squaring both sides to find $\lambda_{Fe}$:
$\dfrac{1}{\lambda_{Fe}} \approx (0.07119)^2 \approx 0.005068$
$\lambda_{Fe} = \dfrac{1}{0.005068} \approx 197.3 \text{ pm}$

The closest value among the given options is $198 \text{ pm}$.

Answer: $198 \text{ pm}$