Question

${}_{80}{}^{197}\mathrm{Hg}$ decay to ${}_{79}{}^{197}\mathrm{Au}$ through electron capture with a decay constant of 0.257 per day. (a) What other particle or particles are emitted in the decay? (b) Assume that the electron is captured from the K shell. Use Moseley's law √v = a(Z − b) with a = 4.95 × 10⁷ s^−1/2 and b = 1 to find the wavelength of the K_α X-ray emitted following the electron capture.

Answer 1

Given:
Decay constant of electron capture = 0.257 per day

(a) The reaction is given as
${}_{80}{}^{197}\mathrm{Hg}+\mathrm{e}\to {}_{79}{}^{197}\mathrm{Au}+\mathrm{v}$
The other particle emitted in this reaction is neutrino v.

(b) Moseley's law is given by
$\sqrt{v}$ = a(Z − b)

We know
$v=\frac{c}{\lambda }$
Here, c = Speed of light
$\lambda$ = Wavelength of the K_α X-ray

$$\begin{gathered} \sqrt{\frac{c}{\lambda }}=4.95\times {10}^7(79-1) \\ =4.95\times {10}^7\times 78 \\ \Rightarrow \frac{c}{\lambda }=(4.95\times 78{)}^2\times {10}^{14} \\ \Rightarrow \lambda =\frac{3\times {10}^8}{149073.21\times {10}^{14}} \\ =20\mathrm{pm} \end{gathered}$$