In a radioactive decay chain reaction, ²³⁰₉₀Th nucleus decays into ²¹⁴₈₄Po nucleus. The ratio of the number of α to number of β⁻ particles emitted in this process is _____.

The correct answer is 2.

Step-by-Step Explanation:

Let us analyze the radioactive decay of Thorium-230 (${\text{Th}}_{90}^{230}$) into Polonium-214 (${\text{Po}}_{84}^{214}$).

Let the number of emitted α particles be $N$_α and the number of emitted β^- particles be $N$_β.

Recall the representations for each emission:
- An alpha particle (α) is a Helium nucleus: ${\text{He}}_2^4$ (Mass number = 4, Atomic number = 2)
- A beta-minus particle (β^-) is an electron: ${\text{e}}_{-1}^0$ (Mass number = 0, Atomic number = -1)

The complete nuclear decay equation can be written as:

${\text{Th}}_{90}^{230}\to {\text{Po}}_{84}^{214}+N_{}$_α(He24)+N_β(e-10)

1. Conservation of Mass Number (A):
The total mass number on the left side must equal the total mass number on the right side.

$230=214+4N_{}$_α+0×N_β

$230-214=4N_{}$_α

$16=4N_{}$_α

$N_{}$_α=4

So, 4 α particles are emitted.

2. Conservation of Atomic Number (Z):
The total atomic number (charge) on the left side must equal the total atomic number on the right side.

$90=84+2N_{}$_α+(-1)×N_β

Substitute $N_{}$_α=4 into the equation:

$90=84+2\left(4\right)-N_{}$_β

$90=84+8-N_{}$_β

$90=92-N_{}$_β

$N_{}$_β=92-90=2

So, 2 β^- particles are emitted.

3. Finding the required ratio:
We need to find the ratio of the number of α particles to the number of β^- particles ($N_{}$_α/N_β):

$\text{Ratio}=\frac{N_{}}{}$_αN_β=42=2

Thus, the ratio of the number of α particles to the number of β^- particles emitted is 2.