Neutron and protons, bff's?

Classical Mechanics Level 4

The bond between protons and neutrons inside the atomic nucleus is formed by the strong interaction. In classical terms, we can think of it as protons and neutrons exchanging a "meson," a kind of fundamental particle between them. Let's estimate the range of the interaction that's caused by exchange of the π \pi meson which has mass of approximately 2.5 × 1 0 − 28 kg 2.5 \times 10^{-28}~\mbox{kg} and decays. To do this, note that from the uncertainty principle one can estimate the lifetime t t of a particle of mass m m by t = ℏ 2 m c 2 t=\frac{\hbar}{2mc^2} , where ℏ = 1.05 × 1 0 − 34 J s \hbar=1.05 \times 10^{-34} ~\mbox{J~s} is known as the (reduced) Planck's constant and c = 3 × 1 0 8 m/s c=3 \times 10^8 ~\mbox{m/s} is the speed of light.

Express your answer in femtometers 1 fm = 1 0 − 15 m 1~\mbox{fm}=10^{-15}~\mbox{m} .

Details and assumptions

  • Nothing can travel faster than the speed of light.


The answer is 0.7.

David Mattingly by David Mattingly

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4 solutions

Mesons travel almost at the speed of light. Hence a fairly accurate estimate of the distance they can travel away from the emitting nucleus before they decay is given by r = c t r= ct where c is their constant speed (which is the speed of light with negligible error) and t the average lifetime of the mesons. Then we have r = h / 2 m c = 0.7 f m r= h/2mc =0.7 fm .

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Armaan Bhullar
May 20, 2014

Simply we calculate life time of a meson and then multiply the resulting value by speed of light and finally convert the answer in femtometers by the given conversion to get the final answer as .7 fm

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David Mattingly Staff
May 13, 2014

Since nothing can exceed the speed limit of the universe -- the speed of light -- the π \pi meson cannot travel further than c c times its lifetime. The maximum range of the force would then be on the order of r = c t = c × ℏ 2 m c 2 = ℏ 2 m c = 0.7 fm r=ct=c \times \frac{\hbar}{2mc^2} = \frac{\hbar}{2mc}= 0.7~\mbox{fm} .

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Frank Aiello
Jan 15, 2018

t t = ℏ 2 m c 2 \frac{\hbar}{2mc^2}

= 1.05 × 1 0 − 34 J − s 2 ( 2.5 × 1 0 − 28 k g ) ( 3 × 1 0 8 m / s ) 2 \frac{1.05 \times 10^{-34} J-s}{2(2.5 \times 10^{-28} kg)(3 \times 10^8 m/s)^2}

= 2.343 × 1 0 − 24 2.343 \times 10^{-24} s s

R s t r o n g R_{strong} = c t ct

= ( 2.343 × 1 0 − 24 s ) (2.343 \times 10^{-24} s) ( 3 × 1 0 8 m / s ) (3 \times 10^{8} m/s)

= 7.029 × 1 0 − 16 7.029 \times 10^{-16} m m

= 0.7029 0.7029 f m fm

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