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Cerium-141 is used to research lung function and it decays by emitting a beta particle. Enter an equation for the beta decay of Ce-141. Express your answer as a chemical equation.

Answer :

To write the equation for the beta decay of Cerium-141 (Ce-141), let's first understand what happens during beta decay:

In beta decay, a neutron in the nucleus is transformed into a proton, and a beta particle (which is essentially an electron) is emitted from the nucleus. This process increases the atomic number by 1 because a neutron changes into a proton, but the mass number remains the same because protons and neutrons have similar mass.

Given:

  • The parent nucleus is Cerium-141 ([tex]^{141}_{58}Ce[/tex])

Beta Decay:

  1. A beta particle ([tex]\beta^-[/tex]) is emitted, which is equivalent to an electron [tex](e^-)[/tex] with a mass number of 0 and an atomic number of -1.

  2. The resulting daughter nucleus will have the same mass number (141) and an atomic number increased by 1 (from Ce with atomic number 58 to Praseodymium (Pr) with atomic number 59).

The Equation for the Beta Decay of Ce-141 is:

[tex]^{141}_{58}\text{Ce} \rightarrow ^{141}_{59}\text{Pr} + \beta^-[/tex]

This chemical equation represents the transformation of Cerium-141 into Praseodymium-141 with the emission of a beta particle. The atomic number of Cerium increases by 1, turning it into Praseodymium, while the mass number stays the same at 141.

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Rewritten by : Barada