Question

In the radioactive decay of 204pb to 200hg the other product is

Answer 1

The transformation of an unstable atomic nucleus into a stable nucleus spontaneously  is known as  radioactive decay. This results in the release of energy or matter from the nucleus of an atom.  The released radiation is in the form of beta particle, alpha particle, gamma rays, etc.

During the radioactive decay, the sum of atomic number and the sum of mass number will be same on both the sides (reactant and product side) of the nuclear decay equation.

Atomic number of $Pb$ is 82 and the atomic number of $Hg$ is 80.

The given decay can be written as:

$_{82}^{204}Pb\rightarrow _{80}^{200}Hg + _{z}^{a}X$

Let X be the other product formed where z is the atomic number of X and a is the mass number.

in order to determine the identity of X:

The value of a:

$204 = 200 + a$

$a = 4$

The value of z:

$82 = 80 + z$

$z = 2$

The element with atomic number 2 is Helium, $He$

The radioactive decay of $^{204}Pb$ to $^{200}Hg$ will be written as:

$_{82}^{204}Pb\rightarrow _{80}^{200}Hg + _{2}^{4}He$

Hence, In the radioactive decay of 204pb to 200hg the other product is $_{2}^{4}He$.