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`A._(7)^(14)N` nucleus, when bombarded by `._(2)^(4)He`, converts into `._(8)^(17)O` accoring to the following reaction. `._(7)^(14)N+._(2)^(4)Herarr._(8)^(17)O+._(1)^(1)H` Atomic masses are: `.^(14)N:14.003u,.^(4)He:4.003u,.^(17)O:16.999u,.^(1)H:1.008u` (a) Find the Q value of the reactions (b) Assume that a `.^(4)He` nucleus collides with a free `.^(14)N` nucleus originally at rest. Calculate the minimum kinetic energy `(k_(0))` that `.^(4)He` must have so as to cause this reaction. (c) If the `.^(17)O` nucleus has an excitation energy of 1.0 MeV, find the minimum kinetic energy `(k_(0))` that `.^(4)He` must have. [Take `1u=930MeV//c^(2)`] |
Answer» Correct Answer - (a) 0.93MeV (b) `k_(0)=1.20MeV` (c) `k_(0)=2.20MeV` |
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