Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
156 views
in Physics by (97.8k points)
closed by
The binding energy per nucleon for `._(6)C^(12) is 7.68 MeV//N` and that for `._(6)C^(13) is 7.47MeV//N`. Calculate the energy required to remove a neutron form `._(6)C^(13)`

1 Answer

0 votes
by (98.2k points)
selected by
 
Best answer
Correct Answer - 4.95 MeV
Total B.E. of `._(6)C^(12)=12xx7.68=92.16 MeV`
Total B.E. of `._(6)C^(13)=13xx7.47=97.11 MeV`
`._(6)C^(13)` has one excess neutron than `._(6)C^(12)`,
`:.` Energy required to remove a neutron
`=97.11-92.16=4.95MeV`

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

...