Solutions for Recitation Exercise # 10
1. (a) In this case, ∆V = Vanode − Vcathode > 0. This means that for (negatively charged) electrons, the
potential energy is lower at A than at C, so electrons coming off the cathode see the anode as being “downhill” from where they start. (Or: It means that the electric field points from A to C, so electrons feel a
force F = −eE towards the anode.) Any electron freed from the cathode will be able to make it across.
(b) Since ∆V < 0, the electrons now see the anode as being “uphill” – electrons have a higher potential
energy there than at the cathode where they start. (Or: The E field now points from cathode to anode,
so electrons feel a force back towards the cathode.) The question now is whether they have enough energy
to make it up the hill. The “stopping potential” is defined by making the hill just high enough that even
the most energetic electrons can’t quite make it. In other words, eVstop = Kmax . How much energy do
the electrons have? They take all of a single photon’s energy hf and then lose at least φ getting free of the
cathode. So eVstop = hf − φ. This corresponds to ∆V = −Vstop (minus because of the electon’s negative
charge – the anode is at a lower potential so that for electrons, its at a higher PE). Since ∆V > −Vstop ,
the most energetic electrons will be able to make it across, but some of the less energetic ones won’t.
(c) Now, since ∆V < −Vstop , not even the most energetic electrons will get across.
2. The work function of a metal measures: (ii) How tightly the electrons are bound within the metal.
3. The photoelectrons from metal 1 will have a lower speed than those from metal 2. Since both are
illuminated with light of the same frequency, the photons hitting both metals will have the same energy.
(i.e. electrons get the same amount of energy when hit by a photon, whether they are in metal 1 or metal
2.) Since metal 1 has a larger work function, more energy is used in freeing an electron, so it will have a
lower kinetic energy, and therefore a lower speed, than it would after leaving metal 2.
4.i Light intensity increased → more photons hitting the metal, so more electrons are released and more
make it to the other side. This means that the current will be greater. The stopping potential won’t change,
since neither the energy of each incoming photon, nor the energy required to remove an electron from
the metal have changed.
ii Light frequency increased → stopping potential increased (eV0 = hf −φ). Therefore, −Vstop moves farther
to the left. The light’s intensity is not changed, so the maximum current won’t change. (Remember that
“frequency” refers to frequency of oscillation of the electric and magnetic field, not rate of photons per
second, even though the units would be right.)
iii Cathode work function increased → stopping potential decreased, so −Vstop increases (moves to the right).
Just like in part (ii), the maximum current is unchanged.
5.i No change – this graph refers only to the energy relationship between frequency of photons and the
potential at which (some) electrons barely have enough energy to reach the anode and form a current. It
is true independent of the number of photons.
ii Use eV0 = hf − φ. For a given frequency (i.e. a given energy that each photon has), if the work
function φ is increased, each electron will leave the metal with less kinetic energy, so Vstop will decrease. So the line will shift to the right. (Why right? For several points on the f axis, with f > f0 ,
draw a vertical line. If Vstop has decreased along each of these, the result is a line that moved to the right.)
A more mathematical way: When plotting V0 versus f , this is the equation for a line, with slope h/e and
intercept −φ/e. So if the work function is increased, the slope will stay the same but the intercept will
decrease. (This line doesn’t continue below the f –axis because the stopping potential can’t be negative.)
So the result is that the line is shifted to the right.
6. Since the frequency is increased, there will be faster photoelectrons (answer ii). By increasing the
frequency, the energy of each photon is increased, so each photoelectron will also have more energy, and
thus will move faster.
7. Since the intensity is increased, there will be more photoelectrons (answer i). By increasing the
intensity, the number of photons is increased, so more electrons will get a photon’s energy and leave the
metal.
8.a The current is zero when ∆V = 0, which corresponds to no energy “hill” at all between the cathode
and anode. This means that there are no photoelectrons at all, i.e. the photons don’t have enough energy
to free any of the electrons. You can verify this to be true, since hc
λ < φ. Current won’t change when the
intensity is doubled.
b Increasing ∆V means that any free photoelectrons have to climb up a potential energy hill to reach the
anode. But, since the light’s frequency and the cathode’s work function haven’t changed, there will still
not be any electrons freed, and the current will still be zero. (The photoelectric effect can’t be described
by Ohm’s Law.)
c Changing the cathode’s work function to 4.3 eV , while having ∆V = 0, will now give us a current.
This is because
hc
1240 eV · nm
=
= 4.96 eV > 4.3 eV
λ
250 nm
In other words, each incoming photon has enough energy to free an electron.
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