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# Find the number of α and β-particles emitted in the process X86222X2862222Rn⟶X84214X2842214Po. - Chemistry

Numerical

Find the number of α and β-particles emitted in the process $\ce{^222_86Rn -> ^214_84Po}$.

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#### Solution

The emission of one α-particle decreases the mass number by 4 whereas the emission of β-particles has no effect on the mass number.

Net decrease in mass number = 222 – 214 = 8. This decrease is only due to α-particle. Hence, number of α-particle emitted = 8/4 = 2

Now, the emission of one α-particle decreases the atomic number by 2 and one β-particle emission increases it by 1.

The net decrease in atomic number = 86 – 84 = 2

The emission of 2 α-particles causes a decrease in the atomic number by 4. However, the actual decrease is only 2. It means atomic number increases by 2. This increase is due to the emission of 2 β-particles. Thus, 2 α and 2 β-particles are emitted.

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#### APPEARS IN

Balbharati Chemistry 11th Standard Maharashtra State Board
Chapter 13 Nuclear Chemistry and Radioactivity
Exercises | Q 3. (R) | Page 203
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