Nuclear reaction 147N- 42He - 178O- 11H can be written as 147N- p 178O-If true enter 1- else enter 0-

The balanced reaction, #160; ^14_7N+ ^4_2He ⟶ ^17_8O+ ^1_1H can be written as follows: ^14_7N(α, p) ^17_8O ...

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Gamma Decay

Nuclear react...

Question

Nuclear reaction $_{7}^{14} N +_{2}^{4} H e ⟶_{8}^{17} O +_{1}^{1} H$ can be written as $_{7}^{14} N (α, p)_{8}^{17} O$ .
If true enter 1, else enter 0.

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