What is the value of A in the following nuclear reaction?
212/84Po→208/82Pb+A/ZX
Looks like z=2
looks like A= 212-208
X looks an alpha particle.
To find the value of element A in the nuclear reaction, we need to balance the mass numbers on both sides.
Starting with the reactant side:
The mass number of Po-212 is 212 (top number of the isotope symbol).
The mass number of Pb-208 is 208.
Now let's balance the equation:
212/84Po → 208/82Pb + A/ZX
To balance the mass numbers, we subtract 208 from both sides:
212 - 208 = A
4 = A
Therefore, the value of A in the reaction is 4.
To determine the value of A in the nuclear reaction 212/84Po → 208/82Pb + A/ZX, we need to use the principle of conservation of mass number and atomic number.
In a nuclear reaction, both the mass number (A) and the atomic number (Z) must be conserved. The mass number represents the total number of protons and neutrons in an atom, while the atomic number corresponds to the number of protons.
In the given reaction, the mass number on the left side is 212 (assuming no neutrinos or electrons are involved). We can deduce this from the isotope notation 212/84Po, where 212 is the mass number and 84 is the atomic number.
On the right side, we have 208/82Pb, which represents lead with a mass number of 208 and an atomic number of 82.
According to the conservation principle, the mass numbers on both sides of the reaction must be equal. Therefore, 212 = 208 + A.
To find the value of A, we subtract 208 from both sides of the equation:
212 - 208 = 208 + A - 208
4 = A
Hence, the value of A in the nuclear reaction 212/84Po → 208/82Pb + A/ZX is 4.