Question

`""_80^197`Hg decay to `""_79^197`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 × 10⁷s^−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`Hg + e → `""_79^197`Au + `"v"`

The other particle emitted in this reaction is neutrino v.

(b) Moseley's law is given by

`sqrtv = a(Z - b)`

We  know
`v = c/lambda`
Here, c = Speed of light
         `lambda` = Wavelength of the K_α X-ray

`sqrt(c/lambda) = 4.95 xx 10^7(79 - 1)`

= `4.95 xx 10^7 xx 78`

⇒ `c/lambda = (4.95 xx 78)^2 xx 10^14`

⇒`lambda = (3 xx 10^8)/(149073.21 xx 10^14)`

 = 20 `"pm"`