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 t

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The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction N 7 16 + He 2 4 &#8594; 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 , &#160; He 2 4 , &#160; H 1 1 and O 8 19 can be taken to be 16 . 006 &#160; u , &#160; 4 . 003 &#160; u , 1 . 008 &#160; u  and 19 . 003 &#160; u , respectively, where 1 &#160; u = 930 &#160; MeV &#160; 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.

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Correct answer: A. A

Given here:  N 7 16 + He 2 4 &#8594; H 1 1 + O 8 19 Energy absorbed in the reaction is  Q = m N + m He - m H - m O &#215; c 2 = 16 . 006 + 4 . 003 - 1 . 008 - 19 . 003 &#215; 930 &#160; MeV = - 1 . 86 &#160; MeV = 1 . 86 &#160; MeV   Let  v  is velocity of alpha particle. Now, the maximum loss of kinetic energy =   1 2 &#215; m &#215; 4 m 5 m &#215; v 2 &#8658; 1 2 m v 2 = 5 4 &#215; Q = 5 4 &#215; 1 . 86 &#160; MeV = 2 . 325 &#160; MeV &#8756; &#160; n = 2 . 33

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