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80^197Hg decay to 79^197Au through elect...

``_80^197Hg` decay to ``_79^197Au` 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 `sqrt v=a(Z-b)`with a `a=4.95xx10^7 s^(-1//2)` and b=1 to find the wavelenghth of the `K_a` x-ray emitted following the electron capature.

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Verified by Experts

The correct Answer is:
A, B

(a) `p + e rarr n = v neutrino `
(b) `sqrt f = a (z-b) `
` sqrt (( c)/( lambda )) = 4.9.5 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) )/(149032 xx 10^(14)) `
` = 2 xx 10^(5) xx 10^(-5)`
` = 2 xx 1m = 20 pm `
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