In the process of nuclear fission, Uranium-235 (²³⁵U) absorbs a neutron (¹⁰n) and undergoes fission, producing several products. The fission of Uranium-235 can yield various different daughter nuclides, but one common reaction produces Barium-139 (¹³⁹Ba) and Krypton-94 (⁹⁴Kr), along with additional neutrons.
The nuclear equation can be balanced by identifying the missing components:
Given the reaction:
\[ ²³⁵_{92}U + ¹^{0}{1}n \rightarrow ¹^{3}{56}Ba + A + B + 3 \cdot ¹^{0}_{1}n \]
We know:
- The atomic number (number of protons) must balance on both sides.
- The mass number must also balance.
From the equation:
- On the left side: \( 235 + 1 = 236 \) (mass number)
- On the right side: \( 139 + A + 3 = 142 + A \)
Setting these equal gives us:
\[ 236 = 142 + A \] thus, \[ A = 236 - 142 = 94\]
For the atomic numbers:
- On the left side: \( 92 + 1 = 93 \)
- On the right side: \( 56 + B + 3 = 59 + B \)
Setting these equal gives us:
\[ 93 = 59 + B \] thus, \[ B = 93 - 59 = 34\]
Finally, we have A as 94 and B as 34.
Thus, using the periodic table:
- A corresponds to Krypton (K), with an atomic mass of \( 94 \).
- B corresponds to Selenium (Se), with an atomic mass of \( 34 \).
C is simply the total of the products, which is another neutron (1 mass number, 0 atomic number).
So, the complete fission process is represented as:
- A: \( ¹^{0}_{1}n \) - Neutron
- B: \( ²⁴⁶_{34}Se \) - Selenium
- C: \( ⁹⁴_{36}Kr \) - Krypton
Final complete answer:
A: \( \text{Krypton (Kr)} \)
B: \( \text{Selenium (Se)} \)
C: \( \text{another neutron (n)} \)
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