InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Consider the `beta^(-)` decay of `._(12)^(27)Mg` `._(12)^(27)Mgrarr._(13)^(27)Al^(*)+._(-1)^(0)e+vecv` 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. |
|
Answer» Correct Answer - (a) 2.0925MeV (b) `KE_(betalt1.0775MeV` (c) 0.0122 `overset(@)A` (d) 1.013 MeV |
|