4. The discovery of X-rays and
electrons
4.1 Gas discharges
19th century: knowledge of charged atoms/molecules
•electrolysis
•discharges of rarefied gases (vacuum).
•near cathode: “glow charge”, “cathode rays” charged molecules?
ether waves? UV light?
various tubes used for
producing cathode rays
http://www.crtsite.com/index.html
Röntgen,1895: as cathode rays impact on glass walls of tube,
other rays are produced, penetrating glass walls and air.
• detection with fluorescent paper.
• no deflection by magnet.
Penetration of thick bodies, absorption dependent on density of matter.
Frau Roentgen’s hand (1895), improved image (Albert von Kollicker,
1896).
• X-rays: ultraviolet light, produced
when cathode rays are stopped.
• 1897 JJ Thomson and Wiechert:
cathode rays: have charge to mass
ratio 2000 times that of Hydrogen ion.
• corpuscular (particle) nature of the
cathode rays: electrons (Stoney).
4.2 X-ray spectra
X-rays: ultraviolet light, produced when cathode rays are stopped.
minimum wavelength of X-rays (maximum
energy) given by:
eVaccelerating  hf max 
hc
min
Continuous x-ray spectrum produced by
electrons impacting on anode known as
“bremsstrahlung”
(see YF Chapter 38.7)
Impacting electrons can also knock out
tightly bound inner electrons of the anode
material. Refilling “holes” with less tightly
bound electrons leads to emission of x-ray
photons, and characteristic sharp spikes
in spectra.
(YF Figures 41.19 & 41.20)
hc
min
 eVaccelerating
Continuous x-ray spectrum from
tungsten anode, as accelerating
voltage increased
X-ray spectrum from Molybdenum
anode showing continuous
spectrum and characteristic xrays superimposed on spectrum.
Spikes do not change with voltage
4.3 Thomson’s experiment
http://www2.kutl.kyushu-u.ac.jp/seminar/MicroWorld1_E/Part1_E/P17_E/electron_E.htm
(YF 27.21)
• accelerating potential V:
E kin
• thus v  2qV / m ;
1 2
 mv  qV
2
• electrons strike screen at end of tube; path of
electrons controlled by applied electric and magnetic
fields;


 
F  qE  qv  B
• condition for a straight line: qvB = qE;
→
v  E / B  2qV / m
→
q
E2

m 2VB 2
• Experiments: single value for q/m, independent of
material of cathode → particles detected (electrons)
are a common constituent of all matter.
Result: e/m = 1.758820174(71)×1011 C/kg
• value of e determined by Millikan
• 2007: e=1.6021892(46)×10-19 C
→ electron mass me=9.11×10-31 kg
hydrogen atom: mH = 1.67×10-27 kg
• Adaptations of Thomson’s apparatus:
mass-spectrometers
JJ Thomson: "Could anything at first sight seem
more impractical than a body which is so small that its
mass is an insignificant fraction of the mass of an
atom of hydrogen? -- which itself is so small that a
crowd of these atoms equal in number to the
population of the whole world would be too small to
have been detected by any means then known to
science.“
(Recording made in 1934. From the soundtrack of the film, Atomic Physics
copyright
©
J.
Arthur
Rank
Organization,
http://www.aip.org/history/electron/jjsound.htm
Ltd.,
1948.)
4.4 Millikan’s oil drop experiment (1909-1913)
(Figure: http://en.wikipedia.org/wiki/Oil-drop_experiment)
• drop diameter around 10-4 mm
• appropriate voltage makes them remain at rest in the field:
mg = q E = q V/d (V, potential difference, E, electric field)
• Result: charges were small multiples of a basic charge e.
Important is precise determination of radii r of
droplets. Thus: Actual experiment:
Initially no applied electric field.
• Force balance between buoyancy and viscous drag
(Stokes law)
4 3
mg  r g  6rv0
3
• density difference Δρ=ρoil-ρair≈ρoil
• η, viscosity of air, v0 terminal velocity (see comment
below).
• Measurement (microscope) of terminal velocity →
radius r.
Now apply electric field E:
4 3
• Force balance:
r g  qE  6rv1
3
q unknown charge of droplet, determined by
measuring new terminal velocity v1
capacitor: E = V/d (d: separation of plates)
• Millikan: measured charges of thousands of
drops were integer multiples of elementary
charge e (experimental deviation 1%).
Comment on terminal speed:
• Consider fall of object through a viscous medium
• Equation of motion (Newton’s second law):
k, drag constant (Stokes drag for sphere
m
dv(t )
 mg  kv(t )
dt
k  6r )
• perform integration (using separation of variables) to compute velocity v(t)
v
•
t
dv
0 k  0 dt;
g v
m
k
k
ln( g  v) |v0   t ;
m
m
k
g v
m  exp  k t ;
 m 
g
mg 
k 
1

exp[

t ] ;
thus v(t ) 

k 
m 
terminal speed is then given by
(limit of t
∞)
mg
vt 
;
k