In the nuclear reaction $_{12} M g^{24} +_{2} H e^{4} =_{0} n^{1} + ?$ The product nucleus is

_{13} A l^{27}

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_{14} S i^{27}

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_{13} A l^{28}

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_{12} M g^{25}

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Solution

The correct option is B $_{14} S i^{27}$
$_{12} M g^{24} +_{2} H e^{4} \to_{0} n^{1} +_{14} S i^{27}$

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Similar questions

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

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

_{2} H e^{4} +_{7} N^{14} \to_{1} H^{1} + X

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

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

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