The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction N716+He24→H11+O819 in a laboratory frame is n (in MeV). Assume…

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

If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.

MARKS Content Team · Accepted Answer

Correct answer: 2.33. Given here: N716+He24→H11+O819 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×4m)/5m×v^2 ⇒1/2mv^2=5/4×Q =5/4×1.86 MeV =2.325 MeV ∴ n=2.33

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