Sarthaks APP
0 votes
in Physics by (38.1k points)

Nuclei with magic no. of proton Z = 2, 8, 20, 28, 50, 52 and magic no. of neutrons N = 2, 8, 20, 28, 50, 82 and 126 are found to be very stable. (i) Verify this by calculating the proton separation energy Sp for 120Sn (Z = 50) and 121Sb = (Z = 51). 

The proton separation energy for a nuclide is the minimum energy required to separate the least tightly bound proton from a nucleus of that nuclide. It is given by 

Sp = (MZ–1, N + MH – MZ,N) c2

Given 119In = 118.9058u, 120Sn = 119.902199u, 121Sb = 120.903824u, 1H = 1.0078252u. 

(ii) What does the existance of magic number indicate?

1 Answer

0 votes
by (63.6k points)
selected by
Best answer

Since SpSn > SpSb, Sn nucleus is more stable than Sb nucleus.

(ii) It indicates shell structure of nucleus similar to the shell structure of an atom. This also explains the peaks in BE/ nucleon curve.

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.