Question 25

Question

Given the following nuclear reaction:
223/87 Fr ----->0/-1 e+A/Z X


Column A
1.
What is the atomic mass of the new atom?:
What is the atomic mass of the new atom?
2.
What is the atomic number of the new atom?:
What is the atomic number of the new atom?
3.
What is the symbol of the new atom?:
What is the symbol of the new atom?
4.
What type of decay is demonstrated here?:
What type of decay is demonstrated here?
Column B
a.223
b.Alpha Decay
c.87
d.Fr
e.Beta Decay
f.222
g.86
h.88
i.Ra
j.Gamma Decay
k.Ac
l.224
m.89

Answer 1

To analyze the given nuclear reaction:

\[ \frac{223}{87} \text{Fr} \rightarrow \frac{0}{-1} e + \frac{A}{Z} X \]

This indicates that Francium-223 (\( \text{Fr} \)) is undergoing beta decay, emitting a beta particle (electron, \( \frac{0}{-1} e \)) and transforming into a new atom denoted as \( \frac{A}{Z} X \).

Step-by-step breakdown of the reaction:

1. Atomic Mass (A): The atomic mass of the original atom is 223 (for Francium). In beta decay, the atomic mass doesn't change because a neutron is transformed into a proton, but the total mass number remains the same. So, the atomic mass of the new atom is: \[ A = 223 \]

2. Atomic Number (Z): The atomic number of Francium (Fr) is 87. During beta decay, the atomic number increases by 1 (due to the conversion of a neutron into a proton). Therefore, the atomic number of the new atom (X) will be: \[ Z = 87 + 1 = 88 \]

3. Symbol of the New Atom: The new element with atomic number 88 is Radium (Ra). Thus, the symbol of the new atom is: \[ X = \text{Ra} \]

4. Type of Decay: Since a beta particle (electron) is emitted, this process is classified as beta decay: \[ \text{Type of decay} = \text{Beta Decay} \]

Matching answers from Column A to Column B:

1. Atomic Mass of the new atom: a. 223
2. Atomic Number of the new atom: h. 88
3. Symbol of the new atom: i. Ra
4. Type of decay demonstrated here: e. Beta Decay

Final Answers:

1. a. 223
2. h. 88
3. i. Ra
4. e. Beta Decay