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For a stable nuclei the...

For a stable nuclei the

A

Biding energy is large

B

Binding energy per nucleons is large

C

Packing fraction is very large

D

Packing fraction is very small

Text Solution

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The correct Answer is:
To determine the correct options for the stability of a nucleus, we need to analyze the concepts of binding energy and packing fraction. ### Step-by-Step Solution: 1. **Understanding Binding Energy**: - Binding energy is the energy required to disassemble a nucleus into its constituent protons and neutrons. A higher binding energy indicates a more stable nucleus. - The binding energy can also be related to the mass defect, which is the difference between the mass of the nucleus and the sum of the individual masses of its protons and neutrons. **Hint**: Remember that a larger binding energy means the nucleus is more stable. 2. **Binding Energy per Nucleon**: - Binding energy per nucleon is calculated by dividing the total binding energy by the number of nucleons (protons + neutrons). This value helps compare the stability of different nuclei. - For stable nuclei, the binding energy per nucleon is generally large, especially for intermediate mass nuclei (like Iron). **Hint**: Look for the trend in binding energy per nucleon across different elements; it peaks for intermediate mass elements. 3. **Understanding Packing Fraction**: - The packing fraction is defined as the ratio of the mass defect to the mass number of the nucleus. It indicates how tightly packed the nucleons are within the nucleus. - A positive packing fraction suggests that the isotopic mass is greater than the mass number, while a negative packing fraction indicates the opposite. **Hint**: A smaller packing fraction (or a negative value) typically indicates a more stable nucleus. 4. **Evaluating the Options**: - **Option A**: Binding energy is large - This is true for stable nuclei. - **Option B**: Binding energy per nucleon is large - This is also true for stable nuclei. - **Option C**: Packing fraction is very large - This is not generally true for stable nuclei. - **Option D**: Packing fraction is very small - This is true for stable nuclei. 5. **Conclusion**: - The correct options for a stable nucleus are **B** (binding energy per nucleon is large) and **D** (packing fraction is very small). ### Final Answer: The correct options for a stable nucleus are **B and D**.
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Knowledge Check

  • Statement: The first man-made atom produced by artifical transmulation was Tc . Explanation: The phenolmenon of converting a stable nuclei into radioactive one is called artifical radioactivity.

    A
    `S` is correct but `E` is wrong.
    B
    `S` is wrong but `E` is correct.
    C
    Both `S` and `E` are correct and `E` is correct explanation of `S`.
    D
    Both `S` and `E` are correct but `E` is not correct explanation of `S`.

  • Statement 1 : Artificial radioisotopes are obtained from stable nuclei whereas it is not so for natural radioisotopes. Statement 2 : Radioisotopes obtained by artifical transmutation do not behave in the same way as the natural radioactive elements

    A
    Statement 1 is true, statement 2 is true , statement 2 is a correct explanation for statement 3
    B
    Statement 1 is true, Statement 2 is true, statement 2 is not a correct explanation for statement 3
    C
    Statement 1 is true, statement 2 is false
    D
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  • Two radioactive nuclei, P and Q, are present in a given sample and both of them decay into a stable nuclei R with different decay constants. At time t = 0, the number of active nuclei of P are 4N_(0) and that of Q are N_(0) . Half-life of P is 1 minute, where as that of Q is 2 minutes. Initially, there are no nuclei of R present in the sample. When the number of nuclei of P and Q become equal, then the number of nuclei of R present in the sample would be

    A
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    B
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    C
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    D
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