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NCERT Exemplar · Q9

Q.Are the nucleons fundamental particles, or do they consist of still smaller parts? One way to find out is to probe a nucleon just as Rutherford probed an atom. What should be the kinetic energy of an electron for it to be able to probe a nucleon? Assume the diameter of a nucleon to be approximately 10−15 m10^{-15}\ \text{m}.

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To probe a nucleon, an electron’s de Broglie wavelength must be comparable to or smaller than the nucleon’s size (~10−15 m10^{-15}\ \text{m}). Using the de Broglie relation and relativistic energy, the required kinetic energy is about 1.24 GeV1.24\ \text{GeV}.

The key idea here is the same one Rutherford used to probe the atom: to “see” a structure, the probe’s wavelength must be smaller than the size of the target. For a nucleon of diameter 10−15 m10^{-15}\ \text{m}, we need an electron with a de Broglie wavelength λ≲10−15 m\lambda \lesssim 10^{-15}\ \text{m}.

Let’s work through this step by step.

  1. The probing condition In any scattering experiment, the resolving power of the probe is limited by its wavelength. To resolve details of size dd, we need λ≤d\lambda \leq d. For a nucleon, d≈10−15 md \approx 10^{-15}\ \text{m}, so we require:

λ≤10−15 m\lambda \leq 10^{-15}\ \text{m}

  1. De Broglie wavelength For a particle of momentum pp, the de Broglie wavelength is:

λ=hp\lambda = \frac{h}{p}

where h=6.626×10−34 J⋅sh = 6.626 \times 10^{-34}\ \text{J·s} is Planck’s constant.

So the required momentum is:

p≥hλ≈6.626×10−3410−15=6.626×10−19 kg⋅m/sp \geq \frac{h}{\lambda} \approx \frac{6.626 \times 10^{-34}}{10^{-15}} = 6.626 \times 10^{-19}\ \text{kg·m/s}

  1. Check if relativistic The rest mass energy of an electron is mec2=0.511 MeV=8.187×10−14 Jm_e c^2 = 0.511\ \text{MeV} = 8.187 \times 10^{-14}\ \text{J}. The momentum we found corresponds to an energy scale we can estimate:

pc=(6.626×10−19)(3×108)≈1.99×10−10 Jpc = (6.626 \times 10^{-19})(3 \times 10^8) \approx 1.99 \times 10^{-10}\ \text{J}

Converting to MeV:

1.99×10−101.602×10−13≈1240 MeV=1.24 GeV\frac{1.99 \times 10^{-10}}{1.602 \times 10^{-13}} \approx 1240\ \text{MeV} = 1.24\ \text{GeV}

Since pc≈1.24 GeVpc \approx 1.24\ \text{GeV} is much larger than mec2=0.511 MeVm_e c^2 = 0.511\ \text{MeV}, the electron is highly relativistic. We must use relativistic energy-momentum relations.

Watch out

A common mistake is to use the non-relativistic kinetic energy formula K=p2/(2m)K = p^2/(2m). For an electron with momentum 6.6×10−19 kg⋅m/s6.6 \times 10^{-19}\ \text{kg·m/s}, that would give K≈2.7×10−10 J≈1.7 GeVK \approx 2.7 \times 10^{-10}\ \text{J} \approx 1.7\ \text{GeV} — which is close but conceptually wrong because the electron is moving at nearly the speed of light. Always check if pc≫mec2pc \gg m_e c^2 before choosing the formula.

  1. Relativistic energy For a relativistic particle, the total energy EE is:

E2=(pc)2+(mec2)2E^2 = (pc)^2 + (m_e c^2)^2

Since pc≫mec2pc \gg m_e c^2, we can approximate: …

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