A nuclear fission is given below A240 → B100 + C140 + Q(energy) Let binding energy per nucleon of nucleus - Physics - Nuclei

Question

A nuclear fission is given below
A240 → B100 + C140 +
Q(energy)
Let binding energy per nucleon of nucleus A, B and C is 7 6 MeV, 8
1 MeV and 8 1 MeV respectively Value of Q is : -
(Approximately)
(1) 20 MeV (2) 220 MeV (3) 120 MeV (4) 240 MeV

Deepak Kumar · Accepted Answer

Dear Student ,A240→B100+C140+Q(energy)Energy released per nucleon = (8
1-7 6)Mev =0 5MevTherefore energy released =0
5×240=120 Mev
So option (3) is correct here
Regards

A nuclear fission is given below A240 → B100 + C140 + Q(energy) Let binding energy per nucleon of nucleus A, B and C is 7.6 MeV, 8.1 MeV and 8.1 MeV respectively. Value of Q is : - (Approximately) (1) 20 MeV (2) 220 MeV (3) 120 MeV (4) 240 MeV

A nuclear fission is given below
A²⁴⁰ → B¹⁰⁰ + C¹⁴⁰ + Q_(energy)
Let binding energy per nucleon of nucleus A, B and C is 7.6 MeV, 8.1 MeV and 8.1 MeV respectively. Value of Q is : - (Approximately)
(1) 20 MeV (2) 220 MeV (3) 120 MeV (4) 240 MeV