Given the following nuclear reaction:

251/98 Cf - 0/0 Y + 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 atomic symbol of the new atom?
4.
What type of decay is demonstrated here?
Column B
a. Alpha Decay
b. 97
c. 98
d. 247
e. 252
f. Es
g. 99
h. Beta Decay
i. 96
j. 251
k. Pu
l. Bk
m. Gamma Decay
n. Cf

1 answer

Let's analyze the given nuclear reaction step by step.

The reaction is: \[ ^{251}{98} \text{Cf} \rightarrow ^{0}{0} \text{Y} + , ^{A}_{Z} \text{X} \]

Step 1: Identify what is happening in the reaction

  1. California-251 (Cf) is undergoing some decay and emitting a particle.
  2. The emitted particle is the 0/0-Y, which is a gamma photon. Gamma radiation does not change the atomic mass or atomic number of the original nucleus. So, the decay is not affecting the mass number or atomic number directly.

Step 2: Balance the reaction

Since a gamma photon doesn't take away mass or charge, we only need to determine what the product X is.

Let’s consider:

  • The atomic mass before decay is 251.

  • The emission has a mass of 0 (gamma photon), so the mass of the new atom \( A \) is still 251.

  • The atomic number of Cf is 98, and you have emitted a neutral particle (gamma), so the atomic number of the new element \( Z \) remains the same. Thus, the atomic number of X is 98.

Filling out the answers:

  1. What is the atomic mass of the new atom?

    • Since the mass number doesn’t change: d. 247
  2. What is the atomic number of the new atom?

    • Since the atomic number doesn’t change after gamma emission: c. 98
  3. What is the atomic symbol of the new atom?

    • CF decays to an isotope with Z = 98 (which is Curium - Cm): f. Es
  4. What type of decay is demonstrated here?

    • Since the reaction shows gamma emission: m. Gamma Decay

Final Answers Summary:

  1. \( A = 251 \) (phase not included in the final answers)
  2. \( Z = 98 \) (cf)
  3. \( \text{Atomic symbol of new atom} = Es \) or not a clear symbol
  4. \( \text{Type of decay} = \) m. Gamma Decay

Please verify the final framework to ensure correctness!