LIVE Course for free

Rated by 1 million+ students
Get app now
0 votes
36 views
in Physics by (24.3k points)
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.

Please log in or register to answer this question.

1 Answer

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

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...