^ 197 _ 80 Hg decay to ^ 197 _ 79 Au through electron capture wit… - Vidyadip

Question

$\text{ }^{197}_{80}\text{Hg}$ decay to $\text{ }^{197}_{79}\text{Au}$ through electron capture with a decay constant of $0.257$ per day.

What other particle or particles are emitted in the decay?
Assume that the electron is captured from the $K$ shell. Use Moseley's law $\sqrt{\text{v}}=\text{a(Z}-\text{b})$ with $\text{a}=4.95\times10^7\text{s}^{-\frac{1}{2}}$ and $b = 1$ to find the wavelength of the $\text{K}_{\alpha} X-$ray emitted following the electron capture.

✓

Answer

$\text{P + e}\rightarrow\text{n + v}$ neutrino $\big[\text{a}\rightarrow4.95\times10^7\text{s}^{-\frac{1}{2}};\text{b}\rightarrow1\big]$
$\sqrt{\text{f}}=\text{a(z}-\text{b})$

$\Rightarrow\sqrt{\frac{\text{c}}{\lambda}}=4.95\times10^7(79-1)=4.95\times10^7\times78$
$\Rightarrow\frac{\text{c}}{\lambda}=(4.95\times78)^2\times10^{14}$
$\Rightarrow\lambda=\frac{3\times10^8}{14903.2\times10^{14}}$
$=2\times10^{-5}\times10^{-6}$
$=2\times10^{-4}\text{m}$
$=20\text{pm}$

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