Question

Hg19780 decay to Au19779 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 Kα X-ray emitted following the electron capture.

Answer 1

Explanation:

It is given:

Electron capture’s decay constant = 0.257 per day

(a) The reaction is shown as

$\mathrm{Hg} 80197+\mathrm{e} \rightarrow \mathrm{v}+\mathrm{Au} 79197$

In this reaction, the emission of another particle takes place, which is neutrino v.

(b) Moseley’s law is shown as v = a(Z − b)

It is known that $v=c \lambda$

Here, c = light’s speed  

$\lambda=K_{a}$ X-ray’s wavelength

$c \lambda=107(79-1) \times 4.95$

$=4.95 \times 78 \times 107$

$\Rightarrow c \lambda=(4.95 \times 78) 2 \times 1014$

$\Rightarrow \lambda=20 \mathrm{pm}$