Questions: 15. Americium -242 is a beta emitter. Write an equation for this nuclear reaction and identify the product nucleus.

Solution Steps

Americium-242 (\(^{242}_{95}\text{Am}\)) is given as the parent nucleus, and it is specified as a beta emitter. In beta decay, a neutron in the nucleus is converted into a proton, and a beta particle (electron) is emitted.

The general equation for beta decay is: \[ ^{A}_{Z}\text{X} \rightarrow ^{A}_{Z+1}\text{Y} + \beta^{-} + \bar{\nu}_e \] where:

• \(^{A}_{Z}\text{X}\) is the parent nucleus,
• \(^{A}_{Z+1}\text{Y}\) is the daughter nucleus,
• \(\beta^{-}\) is the beta particle (electron),
• \(\bar{\nu}_e\) is the antineutrino.

For Americium-242 (\(^{242}_{95}\text{Am}\)): \[ ^{242}_{95}\text{Am} \rightarrow ^{242}_{96}\text{X} + \beta^{-} + \bar{\nu}_e \] Here, the atomic number increases by 1 (from 95 to 96), while the mass number remains the same (242).

The element with atomic number 96 is Curium (Cm). Therefore, the product nucleus is \(^{242}_{96}\text{Cm}\).

Final Answer