Rutherford scattering: impact parameter vs deflection angle, and why my closest approach answer came out off by 2?

7 answers & replies

• [+][–]Arjun Nair12thSample· 2mo ago

  Your reading of the formula is right, and that's the whole physics. Impact parameter $b$ is the perpendicular distance between the nucleus and the original straight line of the alpha. Large $b$: it passes far away, feels a weak repulsion, small $\theta$. $b = 0$: head-on, $\theta = 180°$. Everything in between is a smooth curve.

  For the closest approach: at the turning point all KE has become electrostatic PE. The alpha has charge $2e$, not $e$, and the nucleus has $Ze$. So

  $K = \frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r_0} \Rightarrow r_0 = \frac{2Ze^2}{4\pi\varepsilon_0 K}$

  That's your missing factor of 2. You forgot the alpha's own charge.

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• [+][–]Neel Chatterjee12thSample· 2mo ago

  Adding a picture thought. Imagine a ball rolling past a hill. If it passes right over the top it gets slowed the most, if it passes far from the hill it hardly notices. So smaller impact parameter = closer to the nucleus = more effect... but I always thought 'closer' means it gets deflected less because it only grazes the nucleus, so small $b$ should give small $\theta$? Not sure about this part.

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  • [+][–]Siddharth RaoDropperSample· 2mo ago

    Neel, other way round. The force is repulsive and goes as $1/r^2$, so the closer the alpha comes the harder it is pushed sideways (or back). There's no 'grazing' because the nucleus is a point-like charge, not a solid ball it can brush against. Small $b$ means big $\theta$. In the head-on case the alpha slows to a stop at $r_0$ and then retraces its path, that's the 180° bounce Pooja is asking about. It never touches the nucleus, it just runs out of kinetic energy first.

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    • [+][–]Pooja Iyer12thSample· 2mo agoAuthor

      okay the 'runs out of KE before touching' thing finally clicked. thank you 🙏 So energy conservation gives $r_0$ and force direction gives $\theta$, two different tools?

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      • [+][–]Siddharth RaoDropperSample· 2mo ago

        Yes exactly. Energy conservation tells you how close it gets, the geometry of the repulsion tells you where it goes after. Mixing them up is what costs marks.

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• [+][–]Harsh Gupta12thSample· 2mo ago

  On the electrons: an alpha is roughly 7000 times heavier than an electron. Throwing a cricket ball at a cloud of mosquitoes, the ball doesn't change direction, the mosquitoes do. So electrons can nudge it only a tiny bit. The big kick has to come from the heavy, concentrated positive thing in the centre. Plum pudding atom could never give 180° back-scatter, because the charge is spread out and the field inside is weak.

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• [+][–]Tanvi Deshmukh12thSample· 2mo ago

  My notes table (I'm weird like that!!):

  Observation	Inference
  most pass undeviated	atom is mostly empty space
  few deflected by large angle	positive charge + mass concentrated in a tiny nucleus
  very few return back	head-on, tiny $b$

  NCERT numerical to check yourself: 5.5 MeV alpha on gold ($Z = 79$) gives $r_0 \approx 4\times10^{-14}$ m using the factor of 2 formula. Nucleus is smaller than that, so the alpha really never reaches it.

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