Transcript No Slide Title

A potential difference V is
maintained between the metal
target and the collector cup
Electrons ejected from C
travel to A and G detects
the flow
Apply voltage V between A and C
to slow the ejected electrons down
When potential matches the initial
KE of the electrons, the flow stops
(most energetic electrons stopped)
Kmax= e Vstop
i
Photoelectric Effect
• Vstop does not depend on the intensity of the light
source for a given frequency f
• => classical physics would predict that if we
increase the amplitude of the alternating electric
field, then a larger kick would be given to the
electron?
• => if light is composed of photons, then the
maximum energy that an electron can pick up is that
of a single photon
Adjust V in negative
sense until current
vanishes
Kmax= e Vstop
Independent of intensity!
Measure Vstop as a
function of frequency f
High intensity
All electrons
reach collector
Low intensity
Photoelectric Effect
Vstop
Different metals
f0
f
• In each case, there is a minimum frequency
f0 for the effect to occur
• cannot be explained classically
Photoelectric Effect
• Classical theory:
• oscillating e/m fields in the light cause electrons in
the metal to oscillate
• average KE ~ amplitude2 ~(electric field)2~Intensity
• Quantum theory:
• light composed of energy packets called photons
E=hf =>an electron absorbs one photon and gains
energy hf (this process is independent
of the intensity)
• not expected classically! => increase intensity or
wait longer for electron to absorb enough energy
Photoelectric Effect
Why do electrons stay in metals? Electrical force lowers
the potential energy
PE
e

• Electron needs a minimum amount of energy  to
escape
•  depends on the type of metal => called the work
function
• if an electron absorbs a photon, then (hf - ) is the
amount of energy left over for KE
Photoelectric Effect
• Hence we need hf >  to just escape
• that is f > /h =f0
• Einstein: Kmax = (hf - ) if no other
= e Vstop losses of energy
are involved
• Vstop =(h/e) f - (/e)
Slope =h/e is independent of the metal!
• units: volts is a unit of electrical potential
• eV = (1.6x10-19) volts is a unit of energy
called an electron volt (eV)
• eVstop =h f -  = h( f - f0) = Kmax
Problem
• A satellite in Earth orbit maintains a panel of solar
cells of area 2.60 m2 oriented perpendicular to the
direction of the Sun’s rays. Solar energy arrives at
the rate of 1.39 kW/m2 (energy/area/time)
• (a) at what rate does Solar energy strike the panel?
• rate=energy/time = 1.39(2.60) =3.61 kW
• (b) at what rate are Solar photons absorbed ?
(=550nm)
• each photon carries E=hc/
=(6.63x10-34)(3x108)/(550x10-9)=3.61x10-19 J
• photons/time = (3.61x103)/(3.61x10-19 ) = 1022 /sec
Problem
• (c) how long would it take for a mole of
photons to be absorbed?
• NA = 6.02 x 1023
• time = NA/(number photons/time)
= (6.02 x 1023)/1022 = 60.2 sec
Problem
• Light strikes a sodium surface and causes
photoelectric emission. If Vstop = 5.0 volts and the
work function  is 2.2 eV, what is the wavelength of
the light?
• Ephoton = hf = hc/
• Kmax = Ephoton -  = hc/ -  = e Vstop
•  = (hc)/(e Vstop +  )
• h = 6.63x10-34 J.s = 6.63x10-34 /1.6x10-19 eV.s
= 4.14 x10-15 eV.s
•  = (4.14 x10-15 eV.s)(3x108m/s)/[5.0 eV+2.2 eV]
=170 nm
Momentum
• 1916 Einstein extended the photon idea
• when light interacts with matter, not only energy but
also linear momentum is transferred via photons
• momentum is also transferred in discrete amounts
p=hf/c = h/ photon momentum
•
E=hf = hc/
p=hf/c= h/
• => E = pc
• recall that E2=p2c2 + m2c4 => m=0 massless
• short wavelength photons have more energy and
momentum!
Compton Effect
• 1923 Compton performed an experiment which
supported this idea
• directed a beam of x-rays of wavelength  onto a
carbon target
• x-rays are scattered in different directions

`

 = 71.1 pm (10-12 m)
` has 2 peaks
Compton Scattering
•
•
•
•
Wavelength ` of scattered x-rays has two peaks
these occur at  and  + 
 >0 is the Compton shift
classical physics predicts  =0
• Quantum picture:
• a single photon interacts with electrons in the
target
• light behaves like a ‘particle” of energy E=hf=hc/
and momentum p=h/  => a collision
Compton Scattering
E`=hf `=hc/`

E=hf=hc/
K=mec2(-1)
•
•
•
•
Conservation of energy E = E` + K
=> E` < E => f ` < f
=> ` > 
X-ray momentum p=h/ p`= h/`
electron momentum pe = mev
Compton Scattering
E`=hf `=hc/`

E=hf=hc/
K=mec2(-1)
•
•
•
•
Conservation of energy E = E` + K
=> E` < E => f ` < f
=> ` > 
X-ray momentum p=h/ p`= h/`
electron momentum pe = mev
X-ray scattering
• Energy and momentum are conserved
• Momentum is a vector! F=dp/dt=0 => p = constant