Question:

The disintegration energy Q for the nuclear fission of ²³⁵U → ¹⁴⁰Ce + ⁹⁴Zr + n is _____ MeV.

Given atomic masses:
²³⁵U = 235.0439 u 
¹⁴⁰Ce = 139.9054 u 
⁹⁴Zr = 93.9063 u 
n = 1.0086 u 

Value of c² = 931 MeV/u.

Updated On: Jan 13, 2026
Hide Solution
collegedunia
Verified By Collegedunia

Correct Answer: 208

Approach Solution - 1

1. Calculate Total Mass of Reactants (\(m_r\)):
\[ m_r = 235.0439 \, \text{u}. \]

2. Calculate Total Mass of Products (\(m_p\)):
\[ m_p = 139.9054 + 93.9063 + 1.0086 = 234.8203 \, \text{u}. \] 

3. Calculate Disintegration Energy (\(Q\)):
The disintegration energy \(Q\) is given by:
\[ Q = (m_r - m_p)c^2. \] 

Substitute the values:
\[ Q = (235.0439 - 234.8203) \times 931. \] 

Simplify:
\[ Q = 0.2236 \times 931 = 208.1716 \, \text{MeV}. \] 

Answer: \(208 \, \text{MeV}\)

Was this answer helpful?
0
0
Hide Solution
collegedunia
Verified By Collegedunia

Approach Solution -2

Step 1: Write the given nuclear reaction
The given fission reaction is:
²³⁵U → ¹⁴⁰Ce + ⁹⁴Zr + n
We are required to calculate the disintegration energy (Q-value) released in this process.

Step 2: Recall the formula for Q-value
The Q-value for a nuclear reaction is given by:
\[ Q = \left( m_{\text{parent}} - \sum m_{\text{products}} \right) c^2 \] If the mass of the products is less than the parent mass, the reaction releases energy (exothermic).

Step 3: Substitute the given atomic masses
Parent nucleus mass (Uranium-235): \( m_{\text{U}} = 235.0439\,\text{u} \)
Product masses:
\( m_{\text{Ce}} = 139.9054\,\text{u} \)
\( m_{\text{Zr}} = 93.9063\,\text{u} \)
\( m_n = 1.0086\,\text{u} \)

Step 4: Calculate the total mass of products
\[ m_{\text{products}} = 139.9054 + 93.9063 + 1.0086 = 234.8203\,\text{u} \]

Step 5: Calculate the mass defect
\[ \Delta m = m_{\text{parent}} - m_{\text{products}} \] \[ \Delta m = 235.0439 - 234.8203 = 0.2236\,\text{u} \]

Step 6: Calculate the Q-value (energy released)
\[ Q = \Delta m \times c^2 = 0.2236 \times 931 = 208.1\,\text{MeV} \]

Step 7: Interpretation
This energy corresponds to the amount of nuclear binding energy released during the fission of one uranium-235 nucleus into the given fragments. The large value of 208 MeV explains why nuclear fission is such a powerful source of energy.

Final Answer:
208
Was this answer helpful?
0
0

Questions Asked in JEE Main exam

View More Questions