Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

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

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
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
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon
personalised 1:1 online tutoring
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