Question

provide 4 missing symbols and 2 numbers
232 e 4 0
a → b + c + d
90 88 f 0
a = 1
b = 2
c = 3
d = 4
e = 5
f = 6
a. γ, gamma b. 0 c. 232 d. 90 e. he f. th g. 4
h. 88 i. 188 j. 114 k. re l. -1 m. e n. bi

Explanation:

Step1: Analyze atomic - number and mass - number conservation

In a nuclear decay reaction, the sum of mass - numbers and atomic numbers on both sides of the equation must be equal. The mass - number of the reactant is 232 and atomic number is 90.

Step2: Identify the reactant element

Element with atomic number 90 is Thorium (Th), so $E =$ Th and $A = 232$.

Step3: Determine the product with atomic number 88

Element with atomic number 88 is Radium (Ra). In alpha - decay, an alpha particle ($\alpha$ or $He^{2 + }$) is emitted. An alpha particle has mass - number 4 and atomic number 2. If the product has atomic number 88 and the reactant has atomic number 90, and one of the products is an alpha particle, this is consistent with alpha - decay. So $B$ is an alpha particle, $B=He$.

Step4: Calculate the remaining mass - number and atomic number for other products

The mass - number of the reactant is 232. After emitting an alpha particle (mass - number 4), the remaining mass - number for the other products combined is $232 - 4=228$. The atomic number of the reactant is 90. After emitting an alpha particle (atomic number 2), the remaining atomic number for the other products combined is $90 - 2 = 88$. Since one of the products has atomic number 88, and the other two products ($C$ and $D$) must account for the remaining mass - number and atomic number changes. In many nuclear decay cases, gamma rays ($\gamma$) are emitted which have no mass and no charge. So $C=\gamma$ and $D = 0$.

Step5: Determine the value of $F$

The mass - number of the alpha particle is 4, so $F = 2$.

Answer:

1. C. 232
2. E. He
3. A. $\gamma$, gamma
4. B. 0
5. F. Th
6. G. 4