Given the following nuclear reaction:

Question

223/87 Fr - 0/-1 e + A/Z X
Column A
1.
What is the atomic mass of the new atom?
2.
What is the atomic number of the new atom?
3.
What is the symbol of the new atom?
4.
What type of decay is demonstrated here?
Column B
a. 86
b. Ra
c. Gamma Decay
d. 223
e. Ac
f. 87
g. 89
h. Fr
i. Beta Decay
j. Alpha Decay
k. 224
l. 222
m. 88

Answer 1

To analyze the nuclear reaction given, we can start with the original isotope \( ^{223}_{87} \text{Fr} \) (Francium-223), which undergoes decay in the presence of a beta particle (\( 0^{-1} e \)), also known as a beta decay.

1. Atomic Mass of the New Atom:

   • The atomic mass before decay is 223 (from Francium-223).
   • In a beta decay, a neutron is converted into a proton, which means the atomic mass remains the same (223). However, the atomic number changes because a proton is created.
   • Answer: d. 223

2. Atomic Number of the New Atom:

   • The original atomic number of Francium (Fr) is 87.
   • During beta decay, since a neutron is converted into a proton, the atomic number increases by 1.
   • So, the new atomic number = 87 + 1 = 88.
   • Answer: m. 88

3. Symbol of the New Atom:

   • The element with atomic number 88 is Radium (Ra).
   • Answer: b. Ra

4. Type of Decay Demonstrated Here:

   • The reaction involves emitting a beta particle, indicating a beta decay process.
   • Answer: i. Beta Decay

In summary, here are the answers:

1. 223 (d)
2. 88 (m)
3. Ra (b)
4. Beta Decay (i)