The equation 88Ra226- 86Rn222-2He4 emits which particle-

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Standard XII

Chemistry

Artificial Radioactivity

The equation ...

Question

The equation $_{88} R a^{226} \to_{86} R n^{222} +_{2} H e^{4}$ emits which particle?

β

-decay

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α

-decay

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γ

-decay

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None of the above

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Solution

The correct option is B $α$ -decay
$α$ - decay
When an unstable atomic nucleus emits two protons and two neutrons the radioactive process is known as alpha decay.

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The equation 88Ra226→86Rn222+2He4 emits which particle?

The correct option is B α-decayα - decay When an unstable atomic nucleus emits two protons and two neutrons the radioactive process is known as alpha decay.

The equation $_{88} R a^{226} \to_{86} R n^{222} +_{2} H e^{4}$ emits which particle?

The correct option is B $α$ -decay
$α$ - decay
When an unstable atomic nucleus emits two protons and two neutrons the radioactive process is known as alpha decay.