Question

What is the average binding energy per nucleon for a C-12 nucleus with a mass defect of 0.0993 amu? (1 amu= 1.66 x 10-27 kg; 1 J = 1 kg m2/s2)

1.48 x 10-11 J

1.24 x 10-12 J

4.12 x 10-21 J

8.94 x 1015 J

Answer 1

E = m x c^2
=0.0993* (3*10^8)^2
= 8937000000000000 (D)

Answer 2

Okay, thats what I got again haha.

Answer 3

Thanks Kevin!

Answer 4

XD

Answer 5

u are welcome

Answer 6

oops wait its, 1.24 x 10-12 J

Answer 7

0_O

Answer 8

Response Feedback:
Using the formula, E = mc2, substitute in the value provided. First, convert amu to kg: 0.0993 amu x 1.66 x 10-27 kg /amu = 1.648 x 10-28 kg. Then, substitute into E = mc2: E = (1.648 x 10-28 kg) (3.00 x 108 m/s)2 = 1.484 x 10-11 kg m2/s2 = 1.484 x 10-11 J. In the C-12 nucleus, there are 12 nucleons. To calculate the energy per nucleon, divide the total binding energy by the number of nucleons: (1.484 x 10-11 J)/ 12 = 1.24 x 10-12 J

Answer 9

Owh yeah.. we should convert it too kg

Answer 10

Sorry....

Answer 11

Oops, ow well. Now we know, its fine.

Answer 12

(1 amu= 1.66 x 10-27 kg)