Question

write the nuclear equation for the decay of po - 210 if it undergoes 2 consecutive alpha decays followed by a beta decay followed by another alpha decay. write the nuclear equation for the release of a beta particle by pb - 210.

Explanation:

Step1: Alpha decay of Po - 210

In alpha decay, the mass number decreases by 4 and the atomic number decreases by 2.

$$^{210}_{84}\text{Po} ightarrow^{206}_{82}\text{Pb}+^{4}_{2}\text{He}$$

Step2: Second alpha decay

$$^{206}_{82}\text{Pb} ightarrow^{202}_{80}\text{Hg}+^{4}_{2}\text{He}$$

Step3: Beta decay

In beta decay, the mass number remains the same and the atomic number increases by 1.

$$^{202}_{80}\text{Hg} ightarrow^{202}_{81}\text{Tl}+^{\ \ 0}_{-1}\text{e}$$

Step4: Third alpha decay

$$^{202}_{81}\text{Tl} ightarrow^{198}_{79}\text{Au}+^{4}_{2}\text{He}$$

Answer:

$$^{210}_{84}\text{Po} ightarrow^{206}_{82}\text{Pb}+^{4}_{2}\text{He}$$

$$^{206}_{82}\text{Pb} ightarrow^{202}_{80}\text{Hg}+^{4}_{2}\text{He}$$

$$^{202}_{80}\text{Hg} ightarrow^{202}_{81}\text{Tl}+^{\ \ 0}_{-1}\text{e}$$

$$^{202}_{81}\text{Tl} ightarrow^{198}_{79}\text{Au}+^{4}_{2}\text{He}$$