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Multi-Electron Configurations and Orbital Diagrams

_ 92 U ^ 235...

Question

$_{92} U^{235}$ nucleus absorbs a neutron and disintegrate in $_{54} {X e}^{139}$ , $_{38} {S r}^{94}$ and $X$ so, what will be product $X$ ?

3

-neutrons

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2

-neutrons

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α

-particle

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β

-particle

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Solution

The correct option is C $3$ -neutrons
$_{92} U^{235}$ nucleus absorbs a neutron and then disintegrate into $_{54} {X e}^{139}$ , $_{38} {S r}^{94}$ and $X$ .

Thus,

$_{92} U^{235} +_{0} n^{1} ⟶_{54} {X e}^{139} +_{38} {S r}^{94} + 3_{0} n^{1}$

Hence, option A is correct.

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Similar questions

{_{92} U}^{235}

{_{54} X e}^{139}, {_{38} S r}^{94}

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_{92} U^{235}

_{54} {X e}^{139}

_{38} {S r}^{94}

{_{92} U}^{235}

{_{54}^{92} X e}^{139}, {_{38} S r}^{94}

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_{92} U^{235}

_{54} X e^{139},_{38} S r^{94}

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