According to the nuclear reaction 4 Be -2 He 4-6 C 12-0 n 1- mass number of Be atom isA- 7B- 6C- 9D- 4

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Group Displacement Law

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According to the nuclear reaction $_{4} B e +_{2} H e^{4} \to_{6} C^{12} +_{0} n^{1}$ , mass number of (Be) atom is

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_{4} B e +_{2} H e^{4} \to_{6} C^{12} +_{0} n^{1}

_{4} B e^{9} +_{2} H e^{4} \to_{6} C^{12} +_{0} n^{1}

_{4} B e^{9} +_{2} H e^{4} \to_{6} c^{12}

_{12} M g^{24} +_{2} H e^{4} =_{0} n^{1} + ?

_{12} M g^{24} +_{2} H e^{4} =_{0} n^{1} + ?

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