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Consider the `beta^(-)` decay of `._(12)^(27)Mg`
The atomic masses of `Mg^(27)` and `Al^(27)` are 26.98434u and 26.98154u respectively. Mass of electron is 0.00055u. The `Al^(+)` nucleus has excitively. Mass of electron is 1.015MeV.
(a) Find the Q value of the `beta^(-)` decay.
(b) What is maximum possible kinetic energy of emitted `beta^(-)` particle?
(c) Find the smallest wavelenght photon that `Al^(+)` can emit when it de-excites.
(d) As an alternative to gamma decay, the excited `Al^(+)` nucleus returns to its ground state by giving up its excitation energy to an atomic electron which has a binding energy of 2300eV. The process is known as internal conservation. Find the kinetic energy of the emitted electron.

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1 Answer

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by (24.2k points)
Correct Answer - (a) 2.0925MeV
(b) `KE_(betalt1.0775MeV`
(c) 0.0122 `overset(@)A`
(d) 1.013 MeV

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