Question

The nuclide${\mathbf{}}^{\mathbf{244}}\mathbf{Pu}\mathbf{(}\mathbf{Z}\mathbf{=}\mathbf{94}\mathbf{)}$is an alpha-emitter. Into which of the following nuclides does it decay:${\mathbf{}}^{\mathbf{240}}\mathbf{Np}\mathbf{(}\mathbf{Z}\mathbf{=}\mathbf{93}\mathbf{)}\mathbf{,}{\mathbf{}}^{\mathbf{240}}\mathbf{U}\mathbf{(}\mathbf{Z}\mathbf{=}\mathbf{92}\mathbf{)}\mathbf{,}\mathbf{}{\mathbf{}}^{\mathbf{248}}\mathbf{Cm}\mathbf{(}\mathbf{Z}\mathbf{=}\mathbf{96}\mathbf{)}\mathbf{}\mathbf{or}\mathbf{}{\mathbf{}}^{\mathbf{244}}\mathbf{Am}\mathbf{(}\mathbf{Z}\mathbf{=}\mathbf{95}\mathbf{)}$?

Short answer

The nuclide ${}^{244}\mathrm{Pu}(\mathrm{Z}=94)$will decay into ${}^{240}\mathrm{U}(\mathrm{Z}=92)$through the alpha decay process.

Step-by-step solution

The given data

The nuclide ${}^{244}\mathrm{Pu}(\mathrm{Z}=94)$is an alpha-emitter.

Understanding the concept of alpha decay  

Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle (helium nucleus) and thereby transforms or 'decays' into a different atomic nucleus, with a mass number that is reduced by four and an atomic number that is reduced by two.

Calculation of the product nuclide

Alpha decay of a${}_Z^A$nuclide can be given as:

role="math" localid="1661572189065" ${}_{\mathrm{z}}^{\mathrm{A}}\mathrm{M}\to {}_{\mathrm{z}-2}^{\mathrm{A}-4}\mathrm{N}+{}_2^4\mathrm{He}(\mathrm{\alpha }-\mathrm{particle})$

The nuclide${}^{244}\mathrm{Pu}(\mathrm{Z}=94)$undergoes an alpha decay that can be given using the concept as:${}_{94}^{244}Pu\to {}_{92}^{A-4}\mathrm{N}+{}_2^4\mathrm{He}(\mathrm{\alpha }-\mathrm{particle})$

Hence, plutonium-244 decays into uranium-240 ${}^{240}\mathrm{U}(\mathrm{Z}=92)$.