Question

How many neutrons must an 5626Fe nucleus capture to generate the unstable intermediate [7326Fe] according to the equation: 5626Fe+ ? 10n→[7326Fe]

Answer 1

Answer:

number of neutrons an $_{26}^{56}\textrm{Fe}$ nucleus must capture equal to 17

Explanation:

$_{26}^{56}\textrm{Fe}$ has mass number of 56 and atomic number of 26

$_{26}^{73}\textrm{Fe}$ has mass number of 73 and atomic number of 26

Neutron ($_{0}^{1}\textrm{n}$ has a mass number of 1 and atomic number of 0.

Mass number and atomic number in both side of reaction should be equal.

As atomic number does not change during neutron capture therefore number of neutrons required for the given transformation will be equal to change in mass number of Fe.

So, number of neutrons an $_{26}^{56}\textrm{Fe}$ nucleus must capture = (73-56) = 17