NEET2014PhysicsNuclear PhysicsActual
The Binding energy per nucleon of L i 3 7 and H e 2 4 nucleon are 5.60   M e V and 7.06   M eV , respectively. In the nuclear reaction   3 7 L i +   1 1 H →   2 4 H e +   2 4 H e + Q , the value of energy Q released is
Options
- A19.6 M e V
- B- 2.4 M e V
- C8.4 M e V
- D17.3 M e V
Correct answer
D. 17.3 M e V
Step-by-step solution
L i 7 4   + H 1   1 → H e 4 2   2   B E of products = ( 5.6   M e V × 7 ) + 0 = 39.2   M e V E i = - 39.2   M e V B E of reactant = 7.06 × 4 × 2 = 56.48   M e V E f = - 56.48   M e V As nuclear energy decreases, some energy will be released. Q r e l e a s e = E i - E f = - 39.2 - - 56.48 = 17.28   M e V