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.
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: \(208 \, \text{MeV}\)
Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with
\(K_4\)[Fe(CN)\(_6\)] is : \[ {Cu}^{2+}, \, {Fe}^{3+}, \, {Ba}^{2+}, \, {Ca}^{2+}, \, {NH}_4^+, \, {Mg}^{2+}, \, {Zn}^{2+} \]