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 laborat

Question

The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction ${}_{7}^{16}N + {}_{2}^{4}{He} \to {}_{1}^{1}H + {}_{8}^{19}O$ in a laboratory frame is $n$ (in $MeV$ ). Assume that ${}_{7}^{16}N$ is at rest in the laboratory frame. The masses of ${}_{7}^{16}N, {}_{2}^{4}{He}, {}_{1}^{1}H$ and ${}_{8}^{19}O$ 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

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TestHub · Accepted Answer

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

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