During the bombardment of aluminum with neutrons, particles (nuclei of helium atoms – He) and an isotope of some element were formed. identify this isotope.
13Al27 + 0n1 = 2He4 + 11Na24
In the course of a nuclear reaction, the total charge and mass of particles from the left side of the equation are equal to the charge and mass of particles from the right side.
Taking this into account, we find the charge and mass number of the formed isotope:
Z = Z (Al) + Z (n) – Z (He) = 13 + 0 – 2 = 11 – Na;
A = A (Al) + A (n) – A (He) = 27 + 1 – 4 = 24;
The sought isotope is 11Na24.
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