The meson theory of nuclear forces assumes the virtual exchange ofpions. If a nucleon emits a virtual pion of
rest mass 270 m_e, where m_e is rest mass of electron), the range of nuclear forces is________fm. (Upto two
decimal places)
Correct answer is between '1.40,1.50'. Can you explain this answer?

Question

The meson theory of nuclear forces assumes the virtual exchange ofpions. If a nucleon emits a virtual pion of
rest mass 270 m_e, where m_e is rest mass of electron), the range of nuclear forces is________fm. (Upto two
decimal places)

Correct answer is between '1.40,1.50'. Can you explain this answer?

Answer

Introduction:
The meson theory of nuclear forces proposes that the exchange of pions between nucleons is responsible for the attractive nuclear force. Pions are subatomic particles with a rest mass which is approximately 270 times the rest mass of an electron (me). The range of nuclear forces refers to the distance over which the nuclear force is effective.

Calculation:
To calculate the range of nuclear forces, we can use the uncertainty principle. According to the uncertainty principle, there is a fundamental limit to the precision with which we can simultaneously know the position and momentum of a particle. This uncertainty in momentum can be related to the range of a force.

The uncertainty in momentum (∆p) is given by the formula:
∆p * ∆x ≥ h/2π
where h is the Planck's constant.

Step 1: Calculate the uncertainty in momentum (∆p):
∆p = ∆E / c
where ∆E is the uncertainty in energy and c is the speed of light.

Step 2: Calculate the uncertainty in position (∆x):
∆x = ħ / ∆p
where ħ is the reduced Planck's constant.

Step 3: Calculate the range of nuclear forces (R):
R = ∆x / 2
since the range is defined as the distance over which the nuclear force is effective.

Range Calculation:
In this case, the nucleon emits a virtual pion with a rest mass of 270 me. Since the pion is virtual, it exists only for a very short time and does not violate the conservation of energy.

Step 1: Calculate the uncertainty in energy (∆E):
∆E = mc^2
where m is the rest mass of the pion.

∆E = (270 me) * c^2

Step 2: Calculate the uncertainty in momentum (∆p):
∆p = ∆E / c
∆p = (270 me) * c

Step 3: Calculate the uncertainty in position (∆x):
∆x = ħ / ∆p
∆x = ħ / [(270 me) * c]

Step 4: Calculate the range of nuclear forces (R):
R = ∆x / 2
R = [ħ / (2 * (270 me) * c)]

Final Answer:
By substituting the values of the Planck's constant (ħ), the speed of light (c), and the rest mass of an electron (me) into the above equation, we can calculate the range of nuclear forces. The range falls between 1.40 fm and 1.50 fm when rounded to two decimal places.

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