3B SCIENTIFIC PHYSICS U10700 Mode D'emploi page 6

3. Description
The Planck's constant apparatus is for determining
the magnitude of Planck's constant and the work
done in emitting electrons from a caesium cathode
in a photocell using the back-EMF method.
It contains a vacuum photocell, a voltmeter for
measuring back EMF, a nanoammeter for measuring
the photocell current and a power supply for the
LEDs. Five different light-emitting diodes (LEDs) are
provided, which emit light at differing known aver-
age frequencies. The intensity of the emitted light
can be varied between 0 and 100% in each case. The
photocell itself consists of a cathode with caesium
condensed onto its surface and a ring-shaped anode.
When the apparatus is switched on, a voltage is ap-
plied between the two electrodes and this can be
adjusted by two knobs for coarse and fine adjust-
ment.
Power is supplied to the apparatus via the plug-in
transformer provided. The Planck's constant appara-
tus with order number U10700-115 is designed for a
mains voltage of 115 V (±10%) while the version with
order number U10700-230 is for 230 V (±10%).
4. Technical data
Photocell:
Voltmeter:
Precision:
Nanoammeter:
Precision:
LEDs:
Dimensions:
Weight:
5. Theoretical principles
The light from the LED in the circuit passes through
the ring-shaped anode and hits the cathode, where it
causes emission of electrons with kinetic energy
= ⋅ −
E h f W
kin
According to Albert Einstein's photoelectric theory,
= ⋅
E h f is the energy associated with photons of
light at the wavelength in use.
W is the work done in emitting electrons from the
cathode, i.e. the minimum energy required to expel
electrons from the surface of the metal. This value is
dependent on the material as well as on the tem-
perature. For caesium it is 2.14 eV at 0 K and about 2
eV at room temperature.
Depending on the adjustment of the back EMF be-
tween the cathode and anode, a current of electrons
should flow from the former to the latter. This can
be measured using the nanoammeter. If the back-
EMF corresponds to the critical voltage U
Type 1P39, caesium (Cs)
3½-digit LCD
0.5% (typically)
3½-digit LCD
1% (typically)
472 nm, 505 nm, 525 nm,
588 nm, 611 nm
280x150x130 mm
1.3 kg approx.
where
0,
⋅ = = ⋅ −
e U E h f W
0 kin
then this current should have a magnitude of 0 nA.
e U ⋅ against f for the critical voltages
Plot a graph of
0
U , measured for various frequencies of light f, to
0
obtain a line of gradient h crossing the y axis at –W.
6. Operation
6.1 Measurement of critical voltage at a light
intensity of 75%.
•
Plug in the transformer to supply power.
•
Set the intensity of the light source to 75%.
•
Insert the plug for the first light source into the
LED connector socket.
•
Push together the jaws of the clip for the sleeve
over the collector tube of the photocell and re-
move the sleeve.
•
Push the LED unit fully onto the collector tube of
the photocell until the jaws of the clip snap into
place.
•
Set the fine adjustment knob for the back-EMF
to a central position.
•
Wait a few minutes, then set the photocell cur-
rent roughly to zero using the coarse adjustment
knob.
•
Find an optimum zero point using the fine ad-
justment knob.
•
Take note of the back-EMF as set in this fashion
and record it as the critical voltage U
•
Repeat this measurement for the four other
LEDs.
6.2 Determining Planck's constant h.
•
Work out the frequencies of the light from the
printed wavelengths λ using the formula
c = ⋅
8
where 3 10
•
Use the critical voltages U
e U ⋅ where e=1.6021x10
gies
0
•
Plot the values obtained on a graph of energy
against frequency.
•
Draw a straight line through the points and
determine Planck's constant h from the gradient
and the work W from where the line crosses the
Y axis.
6.3 Proof that the critical voltage does not de-
pend on the light intensity.
•
Select an LED.
•
Set the light to maximum intensity and deter-
mine the critical voltage U
•
Reduce the intensity to zero in a set of steps and
determine the critical voltage U
2
-19
and e=1.6021x10 C,
.
0
m
.
s
to work out the ener-
0
-19
C.
.
0
in each case.
0
c
f =
λ