Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
378 views
in Nuclei by (26.7k points)
closed by

Obtain the binding energy of a nitrogen nucleus 147N Given m = 147N 14.00307 u.

1 Answer

+1 vote
by (26.1k points)
selected by
 
Best answer

Here Z = 7 and A = 14, A – Z = 14 – 7 = 7

∴ Mass defect

= [Z mH + (A – Z)mn – Mn] u

= (7 × 1.00783 + 7 × 1.00867 – 14.00307) u

= (7.05481 + 7.06069 - 14.00307) u = 0.11243 u

Since I u = 931.5 MeV

∴ B.E. of 14N = 0.11243 × 931 Mev

= 104.7 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.

...