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What particle is needed to complete this nuclear reaction?
222 86Rn → 218 84 Po + _____

User Ataddeini
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2 Answers

6 votes

Final answer:

The particle needed to complete the nuclear reaction of radon-222 decaying into polonium-218 is an alpha particle (He). The equation is completed as 22286Rn → 21884Po + 42He, with a significant release of energy if one mole of radon decays.

Step-by-step explanation:

The student has asked about completing a nuclear reaction formula for the decay of radon-222 (Rn). The missing particle needed to complete the nuclear reaction 22286Rn → 21884Po + _____ is an alpha particle, which is represented as 42He or simply He. When radon-222 decays, it emits an alpha particle, thus causing the atomic mass to decrease by 4 units (from 222 to 218) and the atomic number to decrease by 2 units (from 86 to 84), leading to the formation of the daughter isotope polonium-218 (Po). If one mole of radon atoms were to decay, it would release a substantial amount of gamma ray energy, quantified as 4.9 × 107 kJ.

User Darwen
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8.4k points
3 votes

Final answer:

222 86Rn → 218 84 Po + alpha particle (He²⁺ or 4He²). This is because alpha decay involves the emission of an alpha particle, which has 2 protons and 2 neutrons, balancing the nuclear equation.

Step-by-step explanation:

The nuclear reaction presented is a typical alpha decay process. In alpha decay, an alpha particle, which is identical to a helium-4 nucleus consisting of 2 protons and 2 neutrons, is emitted from the nucleus of a radioactive atom. The particle needed to complete the nuclear reaction 22286Rn → 21884Po + _____ is an alpha particle (He²⁺ or 4He²).

To balance the nuclear equation, you would compare the total number of protons and the total number of nucleons (protons plus neutrons) before and after the reaction. Radon-222 has 86 protons and a mass number of 222. After emitting an alpha particle, the resulting Polonium isotope has 84 protons and a mass number of 218.

The alpha particle, therefore, must have 2 protons and a mass number of 4 (4He²) in order to conserve the mass and charge during the decay process.

User Martin James
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8.6k points
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