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प्रश्न
`""_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 Kα X-ray emitted following the electron capture.
उत्तर
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 Kα 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"`
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