The binding energy for the following nuclear reactions are expressed in MeV.
\[ {}^{3}_{2}\text{He} + {}^{1}_{0}\text{n} \rightarrow {}^{4}_{2}\text{He} + 20 \text{ MeV} \] \[ {}^{4}_{2}\text{He} + {}^{1}_{0}\text{n} \rightarrow {}^{5}_{2}\text{He} - 0.9 \text{ MeV} \] If \( X_3, X_4, X_5 \) denote the stability of \( {}^{3}_{2}\text{He}, {}^{4}_{2}\text{He} \) and \( {}^{5}_{2}\text{He} \), respectively, then the correct order is:

Question

A regular hexagon is formed by six wires each of resistance \( r \,\Omega \) and the corners are joined to the centre by wires of same resistance. If the current enters at one corner and leaves at the opposite corner, the equivalent resistance of the hexagon between the two opposite corners will be

Answer 1

To determine the correct order of stability for the nuclei \( {}^{3}_{2}\text{He}, {}^{4}_{2}\text{He} \), and \( {}^{5}_{2}\text{He} \), we need to understand the concept of nuclear binding energy and how it relates to nuclear stability.

Binding energy is the energy required to disassemble a nucleus into its separate protons and neutrons. The higher the binding energy per nucleon, the more stable the nucleus is. Let's analyze the given nuclear reactions:

The first reaction is \( {}^{3}_{2}\text{He} + {}^{1}_{0}\text{n} \rightarrow {}^{4}_{2}\text{He} + 20 \text{ MeV} \). This indicates that when a neutron is added to \( {}^{3}_{2}\text{He} \) to form \( {}^{4}_{2}\text{He} \), 20 MeV of energy is released. This implies that \( {}^{4}_{2}\text{He} \) is more stable than \( {}^{3}_{2}\text{He} \) because forming a more stable nucleus releases energy.
The second reaction is \( {}^{4}_{2}\text{He} + {}^{1}_{0}\text{n} \rightarrow {}^{5}_{2}\text{He} - 0.9 \text{ MeV} \). Here, when a neutron is added to \( {}^{4}_{2}\text{He} \) to form \( {}^{5}_{2}\text{He} \), 0.9 MeV of energy is absorbed. This suggests that \( {}^{5}_{2}\text{He} \) is less stable than \( {}^{4}_{2}\text{He} \), because energy is required (absorbed) to form this nucleus, indicating that it is not as tightly bound.

From these observations, we can conclude:

\( {}^{4}_{2}\text{He} \) is the most stable as it releases a significant amount of energy when formed while other reactions either absorb energy or release less.
\( {}^{5}_{2}\text{He} \) is less stable than \( {}^{4}_{2}\text{He} \) since it requires energy to form, indicating it's not as tightly bound.
\( {}^{3}_{2}\text{He} \) is the least stable in this sequence as \( {}^{4}_{2}\text{He} \) is more stable due to the release of 20 MeV energy.

Thus, the order of stability is: \( X_4 > X_5 > X_3 \).

Answer 2

To determine the correct order of stability for the nuclei \( {}^{3}_{2}\text{He}, {}^{4}_{2}\text{He} \), and \( {}^{5}_{2}\text{He} \), we need to understand the concept of nuclear binding energy and how it relates to nuclear stability.

Binding energy is the energy required to disassemble a nucleus into its separate protons and neutrons. The higher the binding energy per nucleon, the more stable the nucleus is. Let's analyze the given nuclear reactions:

The first reaction is \( {}^{3}_{2}\text{He} + {}^{1}_{0}\text{n} \rightarrow {}^{4}_{2}\text{He} + 20 \text{ MeV} \). This indicates that when a neutron is added to \( {}^{3}_{2}\text{He} \) to form \( {}^{4}_{2}\text{He} \), 20 MeV of energy is released. This implies that \( {}^{4}_{2}\text{He} \) is more stable than \( {}^{3}_{2}\text{He} \) because forming a more stable nucleus releases energy.
The second reaction is \( {}^{4}_{2}\text{He} + {}^{1}_{0}\text{n} \rightarrow {}^{5}_{2}\text{He} - 0.9 \text{ MeV} \). Here, when a neutron is added to \( {}^{4}_{2}\text{He} \) to form \( {}^{5}_{2}\text{He} \), 0.9 MeV of energy is absorbed. This suggests that \( {}^{5}_{2}\text{He} \) is less stable than \( {}^{4}_{2}\text{He} \), because energy is required (absorbed) to form this nucleus, indicating that it is not as tightly bound.

From these observations, we can conclude:

\( {}^{4}_{2}\text{He} \) is the most stable as it releases a significant amount of energy when formed while other reactions either absorb energy or release less.
\( {}^{5}_{2}\text{He} \) is less stable than \( {}^{4}_{2}\text{He} \) since it requires energy to form, indicating it's not as tightly bound.
\( {}^{3}_{2}\text{He} \) is the least stable in this sequence as \( {}^{4}_{2}\text{He} \) is more stable due to the release of 20 MeV energy.

Thus, the order of stability is: \( X_4 > X_5 > X_3 \).

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The Correct Option is A

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