Question

Determine the identity of the nuclide resulting from the electron capture by 55 Fe 56 27 5.Co 55 Mn 25 51 Cr 24 54 Mn 25 55 Co 27

Answer 1

In an electron transfer reaction, the proton rich nucleus of an atom absorbs an inner shell electron. As a result, the charge of the proton is neutralized, and it becomes a neutron. In this process, an electron neutrino (a chargeless entity) is also released.

Since one proton is decreased, the atomic number is decreased by 1, and it becomes a different atom.

This can be written as -

$_{Z}^{A}\textrm{X}+_{-1}^{0}\textrm{e}\rightarrow _{Z-1}^{A}\textrm{Y}+_{0}^{0}\textrm{v}_e$

Where v_e refers to the electron neutrino.

For $_{26}^{55}\textrm{Fe}$ , the reaction will be

$_{26}^{55}\textrm{Fe}+_{-1}^{0}\textrm{e}\rightarrow _{25}^{55}\textrm{Mn}+_{0}^{0}\textrm{v}_e$

So the answer is $_{25}^{55}\textrm{Mn}$ .

So option 2 is correct.

Hope this helps, thanks!