A radioactive nucleus 'A' undergoes a series of decays according to the following scheme:
$A \overset{α}{\to} A_{1} \overset{β}{\to} A_{2} \overset{α}{\to} A_{3} \overset{γ}{\to} A_{4}$
The mass number and atomic number of A are 180 and 72 respectively. What are these numbers for A₄?

Solution

α - decay

, the mass number decreases by 4 and atomic number decreases by 2.

In

β

- decay, mass number remains the same but atomic number increases by 1.

In

γ - decay

, mass number and atomic number remains the same.

Therefore, the equation is,

{}_{72}A^{180} \overset{α}{\to} {}_{70}A_{1}^{176} \overset{β}{\to} {}_{71}A_{2}^{176} \overset{α}{\to} {}_{69}A_{3}^{172} \overset{γ}{\to} {}_{69}A_{4}^{172}

Mass number of A₄ is 172
Atomic number of A₄ is 69

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