One of the new superheavy elements added to the periodic table in 2016 was synthesized when a beam of 48Ca ions was directed at a target of 249Cf. (a) Write a balanced nuclear equation for the formation of this element. (b) What isotope is formed after the nuclide formed in the nuclear transmutation reaction in part (a) emits four alpha particles?

Verified step by step guidance

1. Step 1: Identify the reactants and products in the nuclear reaction. The reactants are 48Ca (Calcium-48) and 249Cf (Californium-249). The product is a new superheavy element, which we can denote as element X with a mass number A and atomic number Z.
2. Step 2: Write the nuclear equation for the synthesis of the new element. The equation is: \[ ^{48}_{20}\text{Ca} + ^{249}_{98}\text{Cf} \rightarrow ^{A}_{Z}\text{X} + ^{1}_{0}\text{n} \] where a neutron is also produced.
3. Step 3: Balance the nuclear equation by ensuring the sum of atomic numbers and mass numbers on both sides are equal. Calculate A and Z for the new element X using the conservation of mass and atomic numbers.
4. Step 4: Determine the isotope formed after the emission of four alpha particles. An alpha particle is \( ^{4}_{2}\text{He} \), so emitting four alpha particles will decrease the mass number by 16 and the atomic number by 8.
5. Step 5: Write the nuclear equation for the decay of the new element by emitting four alpha particles. The equation is: \[ ^{A}_{Z}\text{X} \rightarrow ^{(A-16)}_{(Z-8)}\text{Y} + 4(^{4}_{2}\text{He}) \] where Y is the resulting isotope.