The number of alpha- and beta- particles emit | vedclass

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(A) The radioactive decay process is given by: $_{90}E^{232} \rightarrow _{86}G^{220} + x(\alpha) + y(\beta)$.Number of $\alpha$ particles $(x)$ is ca

Vedclass · Accepted Answer

$3$ $\alpha$ and $2$ $\beta$

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(A) The radioactive decay process is given by: $_{90}E^{232} \rightarrow _{86}G^{220} + x(\alpha) + y(\beta)$.Number of $\alpha$ particles $(x)$ is calculated by the change in mass number: $x = \frac{232 - 220}{4} = \frac{12}{4} = 3$.Number of $\beta$ particles $(y)$ is calculated using the change in atomic number: $90 = 86 + 2x - y$,which gives $y = 86 + 2(3) - 90 = 86 + 6 - 90 = 2$.Therefore,$3$ $\alpha$ particles and $2$ $\beta$ particles are emitted.

The number of $\alpha-$ and $\beta-$ particles emitted when a radioactive element $_{90}E^{232}$ changes into $_{86}G^{220}$ will be

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