The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction N 7 16 + He 2 4 → H 1 1 + O 8 19 in a laboratory frame is n (in MeV ). Assume that N 7 16 is at rest in the laboratory frame. The masses of N 7 16 , He 2 4 , H 1 1 and O 8 19 can be taken to be 16 . 006 u , 4 . 003 u , 1 . 008 u and 19 . 003 u , respectively, where 1 u = 930 MeV c - 2 . The value of n is If the numerical value

Step-by-step solution

Given here: N 7 16 + He 2 4 → H 1 1 + O 8 19 Energy absorbed in the reaction is Q = m N + m He - m H - m O × c 2 = 16 . 006 + 4 . 003 - 1 . 008 - 19 . 003 × 930   MeV = - 1 . 86   MeV = 1 . 86   MeV Let v is velocity of alpha particle. Now, the maximum loss of kinetic energy = 1 2 × m × 4 m 5 m × v 2 ⇒ 1 2 m v 2 = 5 4 × Q = 5 4 × 1 . 86   MeV = 2 . 325   MeV ∴   n = 2 . 33