Thomson's e/m, Millikan's charge, canal rays, neutron: how do the numbers connect, and why was the neutron last?

6 answers & replies

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

  Thomson: a beam of electrons passes through crossed electric and magnetic fields. With only the electric field the beam deflects a certain amount. Now turn on a magnetic field of the right size, in the direction that pushes the beam back, until the spot returns to the undeflected position. At that point $eE = evB$, so you get the speed $v = E/B$. Then remove the electric field. The magnetic field alone bends the beam in a circle of radius $r$, with $evB = mv^2/r$, which gives $e/m = v/(Br) = E/(B^2r)$. All the quantities on the right are measurable, so you get $e/m$ directly. You do need Millikan to separate $e$ from $m$.

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  • [+][–]Rahul DubeyDropperSample· 1mo agoAuthor

    the crossed field logic is what I never saw in the book. Writing out $E/(B^2 r)$ myself now. Thanks all, and Yash, no worries, we all started somewhere.

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• [+][–]Yash Tiwari11thSample· 2mo ago

  cathode rays dependent on which gas is in the tube bhai, if you use hydrogen vs neon then the e/m changes. Thomson tried different gases and electrodes... I think they gave different ratios because lighter gases have lighter particles? Matlab hydrogen ke liye largest e/m aaya tha na, mujhe aisa kuch yaad hai class se.

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  • [+][–]Meera Patel11thSample· 2mo ago

    Yash, that's the exact opposite of what Thomson found. The nature of the cathode rays is the same whatever gas is in the tube and whatever the cathode material: the same $e/m$ every time. That is why they're called a universal constituent of matter. It's the positive rays (canal rays, found by Goldstein) which depend on the gas: each gas gives positive ions of different mass, and the lightest ions (from hydrogen) have the largest $e/m$. The proton is the hydrogen ion, with mass about $1.67\times10^{-27}$ kg, roughly 1836 times the electron's mass.

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• [+][–]Diya Menon11thSample· 2mo ago

  For (3): Millikan sprayed fine oil droplets into the chamber, ionised them with X-rays, and balanced gravity against an electric field to find the charge on each drop. He measured lots of drops and found every charge was a whole-number multiple of one value: $1.6\times10^{-19}$ C. No drop ever had a fractional piece of it. If there was a smaller unit, some drops would have shown it. That is the whole logic. (Quarks have 1/3 charges but they are never free, and that is outside NCERT.)

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• [+][–]Kabir Joshi11thSample· 1mo ago

  Neutron: Chadwick (1932) bombarded beryllium with alpha particles and got a very penetrating radiation. It was not deflected by electric or magnetic fields, so it carried no charge, and from its knocking out protons from paraffin it had a mass nearly equal to a proton.

  $_4^9\text{Be} + {}_2^4\text{He} \rightarrow {}_6^{12}\text{C} + {}_0^1\text{n}$

  Your guess is right: all of the earlier detection tricks rely on charge (deflection in fields, ionisation). No charge, no easy handle, so it took longer.

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