The energy released per fission of the nucleus of \(^{240}\text{X}\) is \(200 \, \text{MeV}\). The energy released if all the atoms in \(120 \, \text{g}\) of pure \(^{240}\text{X}\) undergo fission is _____ \(10^{25} \, \text{MeV}\). (Given \(N_A = 6 \times 10^{23}\))
Correct Answer: 6
Approach Solution - 1
The number of moles of \(^{240}\text{X}\) in \(120 \, \text{g}\) is:
\[ \text{No. of moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{120}{240} = 0.5 \, \text{moles} \]
The number of nuclei in \(0.5\) moles is:
\[ \text{No. of nuclei} = 0.5 \cdot N_A = 0.5 \cdot 6 \times 10^{23} = 3 \times 10^{23} \]
The total energy released is:
\[ E = (\text{No. of nuclei}) \cdot (\text{Energy per fission}) \]
Substitute:
\[ E = (3 \times 10^{23}) \cdot (200) = 6 \times 10^{25} \, \text{MeV} \]
Thus, the total energy released is \(6 \times 10^{25} \, \text{MeV}\).
Approach Solution -2
The correct answer is 6.
No. of mole =240120=21
No. of molecules −21×NA
Energy released =21×6×1023×200
=6×1025MeV
Top Questions on Nuclei
- If the radius of the first Bohr orbit is \( r \), then the radius of the second Bohr orbit will be
- The mass defect in a particular reaction is \( 0.4 \, \text{g} \). The amount of energy liberated is \( n \times 10^7 \, \text{kWh} \), where \( n = \) _____. (speed of light \( = 3 \times 10^8 \, \text{m/s} \))
- The half-lives of two radioactive materials A and B are respectively \( T \) and \( 2T \). If the ratio of the initial masses of the materials A and B is 8:1, then the time after which the ratio of the masses of the materials A and B becomes 4:1 is:
- The disintegration energy \( Q \) for the nuclear fission of \( ^{235}\text{U} \to ^{140}\text{Ce} + ^{94}\text{Zr} + n \) is _____ MeV.
Given atomic masses:
\[^{235}\text{U} = 235.0439 \, u, \quad ^{140}\text{Ce} = 139.9054 \, u, \quad ^{94}\text{Zr} = 93.9063 \, u, \quad n = 1.0086 \, u\]Value of \( c^2 = 931 \, \text{MeV/u} \). - In a radioactive sample \( 2 \times 10^8 \) nuclei reduce to \( 10^8 \) nuclei in 15 minutes, then the half-life of the sample in minutes is:
Questions Asked in JEE Main exam
- A force of 49 N acts tangentially at the highest point of a sphere (solid of mass 20 kg) kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is:
- JEE Main - 2025
- Quantum Mechanics
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
- JEE Main - 2025
- Relations and functions
- If \( \alpha + i\beta \) and \( \gamma + i\delta \) are the roots of the equation \( x^2 - (3-2i)x - (2i-2) = 0 \), \( i = \sqrt{-1} \), then \( \alpha\gamma + \beta\delta \) is equal to:
- JEE Main - 2025
- Complex numbers
- The sum of all local minimum values of the function \( f(x) \) as defined below is:
\[ f(x) = \begin{cases} 1 - 2x & \text{if } x < -1, \\[10pt] \frac{1}{3}(7 + 2|x|) & \text{if } -1 \leq x \leq 2, \\[10pt] \frac{11}{18}(x-4)(x-5) & \text{if } x > 2. \end{cases} \]- JEE Main - 2025
- Functions
- Let \( \langle a_n \rangle \) be a sequence such that \( a_0 = 0 \), \( a_1 = \frac{1}{2} \), and \( 2a_{n+2} = 5a_{n+1} - 3a_n \).n= 0,1,2,3.... Then \( \sum_{k=1}^{100} a_k \) is equal to:
- JEE Main - 2025
- Sequences and Series
Concepts Used:
Nuclei
In the year 1911, Rutherford discovered the atomic nucleus along with his associates. It is already known that every atom is manufactured of positive charge and mass in the form of a nucleus that is concentrated at the center of the atom. More than 99.9% of the mass of an atom is located in the nucleus. Additionally, the size of the atom is of the order of 10-10 m and that of the nucleus is of the order of 10-15 m.
Read More: Nuclei
Following are the terms related to nucleus:
- Atomic Number
- Mass Number
- Nuclear Size
- Nuclear Density
- Atomic Mass Unit