^ 197 _ 80 Hg decays to ^ 197 _ 79 Au via electron capture, with a...

In electron capture, the parent nucleus ^{197}_{80}\text{Hg}$captures an inner shell electron and transforms into the daughter nucleus ^{197}_{79}\text{Au}$. The vacancy created in the $K$ shell is filled by an outer electron, resulting in the emission of a $K_\alpha$ X-ray characteristic of the daughter element. Therefore, the atomic number to be used is $Z = 79$.

Using Moseley's law:

$\sqrt{\nu} = a(Z - b)$

Substituting $a = 4.95 \times 10^7 \text{ s}^{-1/2}$, $b = 1$, and $Z = 79$:

$\sqrt{\nu} = 4.95 \times 10^7 \times (79 - 1)$

$\sqrt{\nu} = 4.95 \times 10^7 \times 78 = 3.861 \times 10^9 \text{ s}^{-1/2}$

Squaring both sides to find the frequency $\nu$:

$\nu = (3.861 \times 10^9)^2 \approx 1.49 \times 10^{19} \text{ Hz}$

The wavelength $\lambda$ of the emitted X-ray is given by:

$\lambda = \dfrac{c}{\nu} = \dfrac{3 \times 10^8}{1.49 \times 10^{19}}$

$\lambda \approx 2.01 \times 10^{-11} \text{ m} = 20.1 \text{ pm}$

This is approximately $20 \text{ pm}$.

Answer: $20 \text{ pm}$