Question:easy

If the nuclear force between a proton and a neutron is attractive, then the distance between them can be:

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Remember the important value $0.8\text{ fm}$. Below this distance the nuclear force is strongly repulsive, while beyond it (up to a few femtometres) the force becomes attractive.
Updated On: Jun 15, 2026
  • $0.12\text{ fm}$
  • $10^{-3}\text{ fm}$
  • $1.1\text{ fm}$
  • $0.3\text{ fm}$
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Recall the character of the nuclear force.
The nuclear force between nucleons is very short ranged and changes character with separation.
Step 2: Note the repulsive core.
At very small separations, roughly below about $0.8\,fm$, the force becomes strongly repulsive, which keeps nucleons from collapsing together.
Step 3: Note the attractive region.
For separations roughly between $0.8\,fm$ and a few femtometres, the force is attractive, which is what binds the nucleus.
Step 4: Note the long-range fall-off.
Beyond a few femtometres the nuclear force becomes negligible.
Step 5: Test the options.
$0.12\,fm$, $10^{-3}\,fm$ and $0.3\,fm$ all lie in the repulsive core (below $0.8\,fm$), so they cannot give attraction. Only $1.1\,fm$ lies in the attractive window.
Step 6: Conclude.
For the force to be attractive the separation must be about $1.1\,fm$, which is option (3).
\[ \boxed{1.1\,fm} \]
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