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Artificial Radioactivity

Indium-110 de...

Question

Indium-110 decays to cadmium-110. The correct reactions for following decay process is:

_{49}^{110} I n ⟶_{48}^{110} C d +_{1}^{1} e

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_{49}^{110} I n ⟶_{48}^{110} C d +_{1}^{- 1} e

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_{49}^{110} I n ⟶_{48}^{110} C d +_{1}^{0} e

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Solution

The correct option is D $_{49}^{110} I n ⟶_{48}^{110} C d +_{1}^{0} e$
For a balanced reaction, mass and charge of both side should be equal

$_{49}^{110} I n ⟶_{48}^{110} C d +_{1}^{0} e$

m : $110$ $110$

e : $49$ $48 + 1 = 49$

Suggest Corrections

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