During spontaneous fission, a nucleus of 25498Cf can produce 11846Pd, four neutrons, and what other nuclide?

52Te - Tellurium

Explanation: Radioactivity occurs when an unstable nucleus of an element decompose into a stable element by emitting particles or forms of eletromagnetic energy. There are types of radioactivity and Spontaneous Fission is one of them.

In Spontaneous Fission, the unstable, also called parent, nucleus splits into two smaller nuclei and emitts one or more neutrons.

The reaction spontaneous fission of Cf is represented below.

25498Cf ⇒ 11846Pd + 4n + X

The characteristics of each element of the equation are:

1) The element Cf has atomic mass 254 and atomic number 98.

2) Element Pd has atomic mass 118 and atomic number 46.

3) Neutron is represented as n, has no atomic number and atomic mass 1.

Given these properties, to determine the other nuclide, we have to find out the nuclide's atomic number, as each element in the periodic table has its own atomic number.

At the left side of the equation, atomic number is 98. At the right side, element X's atomic number is

98 - 46 + 0 = 52

In the periodic table, the element with atomic number 52 is Tellurium, whose symbol is Te.

The other nuclide for the spontaneous fission is Tellurium (Te).