Chapter 39, Problem 66A

Radon-212 $\left({}_{86}^{212}Rn\right)$ is a radioactive gas with a half-life of 24 minutes. Show that when radon-212 undergoes alpha decay, the isotope polonium-208 is formed $\left({}_{84}^{208}Po\right)$.

To determine

Proof for the Polonium due to decay of one alpha particle from Radon.

Explanation of Solution

Introduction:

The atomic number of the element decreases by two units and atomic mass number by four units due to decay of one alpha particle.

The atomic number of ${}_{86}^{212}\text{Ra}$ atom is $86$ and atomic mass number 212. It consist 86 protons in the nucleus and 126 neutrons in the nucleus.

If one alpha particle decays from Radon-212 atom then the remaining number of proton in the nucleus of new element becomes 84 and atomic mass number of new element becomes 208.

The expression for the decay of radon-212 atom by emission of one alpha particle.

${}_{86}^{212}\text{Ra}{\to }_{84}^{208}\text{Po}{+}_2^4\text{He}$

The number of protons in the nucleus of polonium is also 84 and number of neutrons in the nucleus is 124. So, the new element is polonium.

Conclusion:

Thus, the new element due to decay of one alpha particle from radon-212 is polonium-208 because the number of protons in the nucleus of polonium is also 84 and number of neutrons in the nucleus is 124.