Find x in the following nuclear reaction- 2He4 - 13Al27- xP30 - 0n1

The correct option is C 15C _2He^4+_13Al^27→ _15P^30+_on^1 ...

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Group Displacement Law

Find x in t...

Question

Find $x$ in the following nuclear reaction.

$_{2} {H e}^{4} +_{13} {A l}^{27} \to_{x} P^{30} +_{0} n^{1}$

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Solution

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Similar questions

_{13} {A l}^{27} +

\to_{15} P^{30} +_{0} n^{1}

Q

{_{13} A l}^{27} + {_{2} H e}^{4} ⟶ {_{14} S i}^{30} + {_{1} H}^{1} + Q

{_{13} A l}^{27}

26.9815 a m u

{_{14} S i}^{30}

29.9738 a m u

{_{2} H e}^{4}

4.0026 a m u

{_{1} H}^{1}

1.0078 a m u

x

_{7} N^{14} +_{2} {H e}^{4} \to_{x} O^{17} +_{1} H^{1}

_{1} H^{2} +_{1} H^{3} \to_{2} {He}^{4} +_{0} n^{1}

_{1} H^{2} = 2.014,_{1} H^{3} = 3.016

_{2} {He}^{4} = 4.003,_{0} n^{1} = 1.009 amu

_{13} {Al}^{27} +_{2} {He}^{4} \to_{14} {Si}^{30} +_{1} H^{1} + Q

_{13} {Al}^{27} = 26.9815 amu

_{14} {Si}^{30} = 29.9738

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