Explain the J.J. Thomson experiment to determine the specific charge of an electron.

III Long answer questions · Q1

Q. Explain the J.J. Thomson experiment to determine the specific charge of an electron.

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★

11% · 14/127 Questions

✓ Free question

Concept understanding — Thomson e/m Determination

J.J. Thomson's 1897 experiment - regarded as marking the birth of modern physics - measured the specific charge (charge-to-mass ratio) $e/m$ of the cathode-ray particle (the electron) by subjecting a narrow beam of cathode rays to crossed electric and magnetic fields inside a highly evacuated discharge tube. The beam, produced at the cathode and narrowed by a pinhole in the anode, passes between a pair of parallel plates (providing a uniform electric field $E$ ) while the whole tube sits between the poles of a magnet (providing a uniform magnetic field $B$ perpendicular to both the beam and $E$ ); the beam finally strikes a fluorescent screen, where its landing position reveals how much it has been deflected.

The measurement proceeds in stages. First, with both fields switched on, their strengths are tuned until the beam lands exactly back at its original, undeflected position - meaning the electric force $eE$ and magnetic force $ev B$ on the electron exactly cancel, giving the beam's velocity directly as $v=E/B$ , independent of $e/m$ . Second, since the beam is accelerated from cathode to anode through a known potential difference $V$ , energy conservation gives $eV=\tfrac12mv^2$ , so $e/m=v^2/(2V)$ - substituting the velocity found in the first step yields a first value of $e/m$ . Two further, independent methods corroborate this same value: measuring the beam's deflection under the electric field alone (switching the magnetic field off), and measuring the radius of the beam's circular path under the magnetic field alone (switching the electric field off, so the magnetic force supplies the centripetal force for circular motion).

All three methods consistently give $e/m\approx1.7\times10^{11}\ \text{C kg}^{-1}$ , and - crucially - this value was found to be completely independent of which gas filled the tube and which metal the electrodes were made of, strong evidence that the electron is a truly universal constituent of all matter, not a species-specific byproduct of any one particular gas or metal.

Step 1. Apparatus. A highly evacuated discharge tube produces a narrow cathode-ray beam (selected by a pinhole in the anode disc A), which passes between a pair of parallel deflecting plates (electric field $E$ ) while the whole tube sits between the poles of a magnet (magnetic field $B$ , perpendicular to $E$ and to the beam). The beam finally strikes a zinc-sulphide-coated fluorescent screen, producing a visible scintillation spot.

Step 2. Velocity selection. With both fields switched on, their strengths are tuned until the beam lands at its original, undeflected position O - meaning the electric force and magnetic force on the electron exactly balance: $eE=ev B$ , giving the beam's velocity directly as $v=E/B$ , independent of the electron's mass or charge individually.

Step 3. Specific charge from accelerating energy. The beam is accelerated from cathode to anode through a known potential difference $V$ , so its potential energy converts entirely to kinetic energy: $eV=\tfrac12mv^2$ , giving $\dfrac{e}{m}=\dfrac{v^2}{2V}=\dfrac{1}{2V}\left(\dfrac{E}{B}\right)^2$ .

Step 4. Cross-checks. Two further, independent measurements corroborate this value: (i) with the magnetic field off, the electric-field-only deflection $y$ on the screen (over plate length $l$ ) gives $\dfrac{e}{m}=\dfrac{yE}{Cl^2B^2}$ for a geometric constant $C$ ;

(ii) with the electric field off, the magnetic-field-only circular path of radius $R$ gives $\dfrac{e}{m}=\dfrac{E}{B^2R}$ .

Step 5. Result. All three methods consistently give $e/m\approx1.7\times10^{11}\ \text{C kg}^{-1}$ , and this value was found to be independent of both the gas filling the tube and the metal used for the electrodes - strong evidence that the electron is a universal constituent of all matter.

✓Final answer

e/m\approx1.7\times10^{11}\ \text{C kg}^{-1}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2018Set ANNUAL1 markMCQ
$(e/m)$
(a) $E^2$
(b) $E$
(c) $E^{-2}$
(d) $E^{-1}$
›Reveal solutionSolution

The deflection of the electron beam, with only the electric field $E$ applied, is directly proportional to $E$ .

In Thomson's $e/m$ experiment, when a transverse electric field of intensity $E$ alone is applied between the deflecting plates, each electron in the beam experiences a constant transverse electric force $F = eE$ , producing a transverse acceleration $a = \dfrac{eE}{m}$ .

If the electrons enter with a fixed horizontal speed $v$ and traverse plates of length $l$ , the time spent between the plates is $t = \dfrac{l}{v}$ . The vertical (transverse) deflection acquired while crossing the plates is:

$y = \dfrac12 a t^2 = \dfrac12 \cdot \dfrac{eE}{m}\cdot\left(\dfrac{l}{v}\right)^2$

For a fixed beam speed $v$ and fixed plate geometry $l$ , this shows $y \propto E$ — the deflection grows linearly with the applied electric field. This proportionality is exactly what Thomson used, together with the magnetic-field balancing condition, to extract the charge-to-mass ratio $e/m$ of the electron.

✓Final answer
$E$
CBSE 2017Set ANNUAL1 markMCQ
$Q. Arrange electron (e), proton (p) and deutron$
(d) in the increasing order of their specific charge :
(a) e, p, d
(b) d, p, e
(c) p, e, d
(d) d, e, p
›Reveal solutionSolution

Specific charge $q/m$ is largest for the lightest particle (electron) and smallest for the heaviest (deuteron), giving the increasing order deuteron, proton, electron.

Specific charge (charge-to-mass ratio) is defined as $\dfrac{q}{m}$ , and it is exactly this ratio that a Bainbridge-type mass spectrometer or a Thomson $e/m$ apparatus measures via the radius of circular deflection in a magnetic field ( $r = \dfrac{mv}{qB}$ , so $\dfrac{q}{m}=\dfrac{v}{rB}$ ).

For the electron, the charge magnitude is the elementary charge $e = 1.6\times10^{-19}$ C, and its mass is extremely small, $m_e = 9.11\times10^{-31}$ kg, giving a specific charge $\dfrac{e}{m_e}\approx 1.76\times10^{11}$ C/kg — by far the largest of the three, because the electron is thousands of times lighter than the proton for the same magnitude of charge.

For the proton, the charge is also $+e$ , but the mass is $m_p = 1.67\times10^{-27}$ kg (about 1836 times the electron's mass), giving $\dfrac{e}{m_p}\approx 9.58\times10^{7}$ C/kg.

A deuteron (the nucleus of deuterium, $^2_1\text{H}$ ) consists of one proton and one neutron bound together, so it carries the same single elementary charge $+e$ as the proton, but its mass is roughly double the proton's mass ( $m_d\approx 2m_p$ , since the neutron's mass is close to the proton's). Because the charge is unchanged but the mass is doubled, the deuteron's specific charge is roughly half the proton's: $\dfrac{e}{m_d}\approx 4.79\times10^{7}$ C/kg — the smallest of the three.

Putting the numbers in increasing order of specific charge: deuteron ( $4.79\times10^7$ ) $<$ proton ( $9.58\times10^7$ ) $<$ electron ( $1.76\times10^{11}$ ), i.e. the sequence d, p, e. This directly rules out options (a), (c), and (d), which place the particles in a different order.

✓Final answer
$q/m$

←Previous question · evaluationQ27 — What are the constituent particles of a neutron and a proton?→Next question · evaluationQ2 — Discuss Millikan's oil drop experiment to determine the charge of an e…

Q.Explain the J.J. Thomson experiment to determine the specific charge of an electron.

🎓Unlock everything free for 14 days

Unlock everything free for 14 days