Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9
AI Mentor
Check Your IQ
Free Expert Demo
Try Test

Courses

Dropper NEET CourseDropper JEE CourseClass - 12 NEET CourseClass - 12 JEE CourseClass - 11 NEET CourseClass - 11 JEE CourseClass - 10 Foundation NEET CourseClass - 10 Foundation JEE CourseClass - 10 CBSE CourseClass - 9 Foundation NEET CourseClass - 9 Foundation JEE CourseClass -9 CBSE CourseClass - 8 CBSE CourseClass - 7 CBSE CourseClass - 6 CBSE Course
Offline Centres

Q.

The binding energy per nucleon for C12 is 7.68 MeV and that for C13 is 7.47 MeV. The energy required to remove a neutron from C13 is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

a

0.21 MeV

b

2.52 MeV

c

4.95 MeV

d

2.75 MeV

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

C13→C12+n 1 0 For C13→BE1=13(7.47)=97.11For C12→BE2=12(7.68)=92.16Energy required = 97.11 – 92.16 = 4.95 Mev
Watch 3-min video & get full concept clarity

courses

No courses found

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE
whats app icon