Consider the nuclear equation below.
[tex]\[ _{92}^{235} U \longrightarrow X + _2^4 He \][/tex]

What is the nuclide symbol of [tex]\( X \)[/tex]?

A. [tex]\( _{94}^{231} Pu \)[/tex]

B. [tex]\( _{90}^{235} Th \)[/tex]

C. [tex]\( _{94}^{239} Pu \)[/tex]

D. [tex]\( _{90}^{231} Th \)[/tex]



Answer :

Sure! Let's analyze the given nuclear reaction step by step:

Given the nuclear equation:
[tex]\[ {}_{92}^{235} U \longrightarrow X + {}_{2}^{4} He \][/tex]

1. Identify the given nuclei:
- Uranium ([tex]\( U \)[/tex]) has an atomic number of 92 and a mass number of 235.
- Helium ([tex]\( He \)[/tex]) (often referred to as an alpha particle) has an atomic number of 2 and a mass number of 4.

2. Balance the atomic and mass numbers:
- For any nuclear reaction, the sum of the atomic numbers and the sum of the mass numbers on both sides of the equation must be equal.

3. Calculate the atomic number of [tex]\( X \)[/tex]:
- The total atomic number on the left side (Uranium) is 92.
- On the right side, we have Helium’s atomic number, which is 2.
- Let [tex]\( Z_X \)[/tex] be the atomic number of [tex]\( X \)[/tex].

So, balancing the atomic numbers:
[tex]\[ 92 = Z_X + 2 \][/tex]
[tex]\[ Z_X = 92 - 2 \][/tex]
[tex]\[ Z_X = 90 \][/tex]

4. Calculate the mass number of [tex]\( X \)[/tex]:
- The mass number on the left side (Uranium) is 235.
- On the right side, we have Helium’s mass number, which is 4.
- Let [tex]\( A_X \)[/tex] be the mass number of [tex]\( X \)[/tex].

So, balancing the mass numbers:
[tex]\[ 235 = A_X + 4 \][/tex]
[tex]\[ A_X = 235 - 4 \][/tex]
[tex]\[ A_X = 231 \][/tex]

So, the nuclide [tex]\( X \)[/tex] has an atomic number of 90 and a mass number of 231.

5. Identify the element with atomic number 90:
- The element with atomic number 90 is Thorium ([tex]\( Th \)[/tex]).

Hence, the nuclide symbol of [tex]\( X \)[/tex] is [tex]\( {}_{90}^{231} Th \)[/tex].

So, the correct answer is:
[tex]\[ {}_{90}^{231} Th \][/tex]