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e) $_{95}^{241}am ightarrow _____+_{93}^{237}np$

Question

e) $_{95}^{241}am
ightarrow _____+_{93}^{237}np$

Explanation:

Step1: Apply mass - number conservation

The mass - number of Americium ($Am$) is 241 and of Neptunium ($Np$) is 237. So the mass - number of the missing particle is $241 - 237=4$.

Step2: Apply atomic - number conservation

The atomic - number of $Am$ is 95 and of $Np$ is 93. So the atomic - number of the missing particle is $95 - 93 = 2$.

Step3: Identify the particle

A particle with mass - number 4 and atomic - number 2 is an alpha particle ($\alpha$) or $_2^4He$.

Answer:

$_{2}^{4}He$