Waves and Modern Physics 203-SN3-RE • CEGEP, Montreal

Revision sheet: modern physics (203-SN3-RE)

Modern physics gathers everything that mechanics and electromagnetism could not explain, in Waves and Modern Physics, 203-SN3-RE (formerly 203-NYC): the spectrum of a hot body, the photoelectric effect, the lines of hydrogen, the stability of nuclei. Each phenomenon has its own short relation and its own particular unit.

This sheet gathers the mistakes that cost the most on tests: intensity confused with photon energy, the threshold treated as an obstacle that can be crossed, the negative Bohr levels, Stefan's fourth-power law, and the two conversions of the chapter, the electron-volt and the atomic mass unit. It is written for CEGEP students in the Natural Science program (Sciences de la nature).

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

Intensity and energy stop going together: more light means MORE photons, never more energetic photons. Almost all the traps of the chapter follow from this separation, and the rest from two conversions, the electron-volt and the atomic mass unit.

This chapter is part of 203-SN3-RE, Waves and Modern Physics

The essentials

The photon and the photoelectric effect

• $E=hf=\dfrac{hc}{\lambda}$ with $h=6.63\times 10^{-34}$ J·s, and $1$ eV $=1.6\times 10^{-19}$ J.
• A shortcut to know by heart: $E$ in electron-volts $=\dfrac{1240}{\lambda\text{ in nm}}$ . Visible light, from $400$ to $700$ nm, therefore corresponds to $3.1$ down to $1.8$ eV.
• Photoelectric effect: $E_{photon}=W_{0}+E_{k,max}$ , where $W_{0}$ is the work function of the metal.
• Threshold frequency: $hf_{0}=W_{0}$ . Below it, NO electron is emitted, whatever the intensity and whatever the duration.
• Stopping potential: $eV_{s}=E_{k,max}$ . The light intensity changes the NUMBER of electrons emitted, never the energy of each.

Only the solid part can be measured: below the threshold, no electron is emitted. The dashed extension cuts the vertical axis at

-W_{0}

, and the slope is

h

The graph of $E_{k,max}$ against frequency is a STRAIGHT LINE whose slope is $h$ and whose intercept is $-W_{0}$ : this is the historical measurement of Planck's constant, and a recurring exam question.

Bohr, de Broglie and the blackbody

• Hydrogen atom: $E_{n}=-\dfrac{13.6}{n^{2}}$ eV, all NEGATIVE, with $E_{1}=-13.6$ eV. The ionization energy from the ground state is therefore $13.6$ eV.
• Photon emitted or absorbed in a jump: $E_{photon}=|E_{n_{f}}-E_{n_{i}}|$ . The $3\to 2$ transition gives $1.89$ eV, that is $656$ nm, the red line of hydrogen.
• de Broglie: every particle of momentum $p$ has the wavelength $\lambda=\dfrac{h}{p}$ , with $p=mv$ and, starting from the kinetic energy, $p=\sqrt{2mE_{k}}$ .
• Blackbody, Wien's law: $\lambda_{max}T=2.898\times 10^{-3}$ m·K. Stefan-Boltzmann law: $\dfrac{P}{A}=\sigma T^{4}$ with $\sigma=5.67\times 10^{-8}$ W/(m $^{2}$ ·K $^{4}$ ).
• Doubling the temperature divides $\lambda_{max}$ by $2$ and multiplies the radiated power by $16$ .

The levels, all negative, crowd together toward

E=0

; the

3\to 2

drop emits

1.89

eV, that is

656

nm, and

13.6

eV is needed to ionize the atom from

n=1

The temperatures in Wien's and Stefan's laws are always in KELVINS. A calculation carried out in degrees Celsius gives a power that is wrong by a factor that can reach several thousand.

The Compton effect: the photon has momentum

• A photon carries a momentum $p=\dfrac{h}{\lambda}=\dfrac{E}{c}$ : scattered by a free electron, it gives up part of it and leaves with a LONGER wavelength.
• Compton shift: $\Delta\lambda=\lambda'-\lambda=\dfrac{h}{m_{e}c}(1-\cos\theta)$ , with $\dfrac{h}{m_{e}c}=2.43$ pm and $\theta$ the scattering angle of the photon.
• The shift is zero straight ahead, equals $2.43$ pm at 90° and at most $4.86$ pm at 180°: it depends on the angle, never on the incident wavelength.

The shift depends only on the angle: zero straight ahead,

2.43

pm at 90°, maximal at 180° with

4.86

pm.

The nucleus and relativity

• Decay: $N(t)=N_{0}\left(\dfrac{1}{2}\right)^{t/T}=N_{0}e^{-\lambda t}$ with $\lambda=\dfrac{\ln 2}{T}$ . The activity in becquerels follows the same law.
• Mass defect: $\Delta m$ is the difference between the sum of the masses of the SEPARATED nucleons and the mass of the nucleus. Binding energy: $E=\Delta m\,c^{2}$ , with $1$ u $=931.5$ MeV/ $c^{2}$ .
• The binding energy per nucleon curve peaks at iron-56: fusion releases energy to the left of the peak, fission to the right.
• Relativity: $\gamma=\dfrac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ , dilation $\Delta t=\gamma\Delta t_{proper}$ , contraction $L=\dfrac{L_{proper}}{\gamma}$ , energy $E=\gamma mc^{2}$ and $K=(\gamma-1)mc^{2}$ .
• General relation: $E^{2}=(pc)^{2}+(mc^{2})^{2}$ , which gives $E=pc$ for the photon, whose mass is zero.

The curve peaks at iron-56 near

8.8

MeV: fusing two light nuclei climbs toward the peak, splitting a heavy nucleus does too, and both release energy.

Time DILATES and lengths CONTRACT: $\gamma$ multiplies one and divides the other. Checking this direction before concluding avoids the most frequent mistake of the relativity part.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Expecting faster electrons from more intense light

the whole question, 6 points

What not to write

“We double the intensity of the lamp, so the electrons emitted have twice the energy and the stopping potential doubles.”

What to write

“The intensity doubles the NUMBER of photons, hence the number of electrons emitted, hence the current. The energy of each electron depends only on the FREQUENCY: $E_{k,max}$ and the stopping potential are unchanged.”

The two green curves saturate at different currents, hence at different NUMBERS of electrons, but they cancel at the SAME

-V_{s}

. The red one, which would double

V_{s}

, is the wrong copy.

Why: This is the heart of Einstein's argument: an electron absorbs ONE photon, not a fraction of the beam. It is this independence between intensity and energy that classical physics could not explain.

2. Believing the threshold can be crossed by waiting or by shining harder

the whole question, 5 points

What not to write

“The frequency is below the threshold, but with a very powerful lamp or a long exposure, the electrons will eventually come out.”

What to write

“Below $f_{0}$ , NO electron is ever emitted: each photon carries less than $W_{0}$ and an electron absorbs only one. Neither the intensity nor the duration changes anything.”

A 600 nm photon (

2.07

eV) does not reach the exit, and two photons never stack on one electron (red); the 400 nm photon (

3.10

eV) clears it alone (green).

Why: Energy accumulating over time is precisely what classical theory predicted, and what experiment refutes: emission is instantaneous above the threshold, and strictly zero below it.

3. Mixing electron-volts and joules

the whole question, and a result faster than light

What not to write

“ $E_{k,max}=0.80$ eV, so $v=\sqrt{\dfrac{2(0.80)}{9.11\times 10^{-31}}}=1.3\times 10^{15}$ m/s.”

What to write

“We must convert: $0.80$ eV $=0.80\times 1.6\times 10^{-19}=1.28\times 10^{-19}$ J, hence $v=5.3\times 10^{5}$ m/s.”

Why: The electron-volt is convenient for comparing energies, but no mechanical formula accepts it. The check is immediate: a speed above $3\times 10^{8}$ m/s is impossible.

4. Subtracting the Bohr levels the wrong way

3 points, and a negative wavelength

What not to write

“Transition from $n=3$ to $n=2$ : $E=-\dfrac{13.6}{9}+\dfrac{13.6}{4}$ ... I find $-1.89$ eV, so a photon of negative energy.”

What to write

“The photon energy is an ABSOLUTE value: $E=|E_{2}-E_{3}|=13.6\left(\dfrac{1}{4}-\dfrac{1}{9}\right)=1.89$ eV, so $\lambda=\dfrac{1240}{1.89}=656$ nm, the red line of hydrogen.”

Why: The levels are negative because the electron is BOUND, and it is their gap that counts. A drop emits, a rise absorbs, and in both cases the photon carries the same positive amount.

5. Making the radiated power double when the temperature doubles

3 points, and a factor of 8 on the result

What not to write

“The star is twice as hot, so it radiates twice as much.”

What to write

“Stefan-Boltzmann goes as $T^{4}$ : doubling the temperature multiplies the power by $2^{4}=16$ . Its colour, for its part, shifts by a factor of $2$ toward the blue, by Wien's law.”

2T

, the area under the curve, the radiated power, is 16 times larger and the maximum moves from 966 to 483 nm; the red curve, simply doubled, is the wrong copy.

Why: The two blackbody laws do not vary at the same rate: one goes as $\dfrac{1}{T}$ , the other as $T^{4}$ . This is what makes hot stars both blue and overwhelmingly powerful.

6. Believing that two half-lives make everything disappear

the whole dating question, 5 points

What not to write

“After two half-lives, nothing is left: $50\,\%$ and then another $50\,\%$ .”

What to write

“Each half-life halves what REMAINS: after two half-lives, $\dfrac{1}{4}$ is left, after three $\dfrac{1}{8}$ . The decay is exponential and never reaches zero.”

After two half-lives, a QUARTER of the nuclei remains (green), not zero (red): each half-life halves what remains, and the curve never reaches the axis.

Why: A percentage applied to what remains does not add up. It is the same reasoning as for compound interest, and it is what makes it possible to date objects far older than the half-life of carbon-14.

7. Calculating a mass defect with the mass of the nucleus alone

the whole question, a factor of 130

What not to write

“The helium-4 nucleus weighs $4.00151$ u, so its binding energy is $4.00151\times 931.5=3728$ MeV.”

What to write

“The mass defect is the DIFFERENCE between the sum of the separated nucleons, $2(1.00728)+2(1.00866)=4.03188$ u, and the mass of the nucleus: $\Delta m=0.03037$ u, that is $E=28.3$ MeV, or $7.07$ MeV per nucleon.”

The mass defect is the GAP between the separated nucleons and the nucleus:

0.03037

u, that is

28.3

MeV to supply to separate everything, and not the whole mass of the nucleus.

Why: The binding energy is what must be SUPPLIED to separate the nucleons, not the total mass energy of the nucleus. The expected order of magnitude is a few MeV per nucleon, never thousands.

8. Expecting a visible Compton effect in visible light

3 points on the discussion question

What not to write

“The Compton shift depends on the wavelength, so it is larger for visible light than for X-rays.”

What to write

“ $\Delta\lambda=\dfrac{h}{m_{e}c}(1-\cos\theta)$ is an ABSOLUTE shift, at most $4.86$ pm. On $550$ nm, that represents one millionth; on $0.071$ nm of X-rays, more than $3\,\%$ : that is measurable.”

The same shift of

4.86

pm clearly separates the two X-ray lines, nearly 7%, and stays invisible at

550

nm, where it weighs only one millionth.

Why: The shift depends only on the angle, never on the incident wavelength. This is precisely why Compton had to work with X-rays to make the effect observable.

9. Swapping time dilation and length contraction

the whole question, 4 points

What not to write

“At high speed, everything shrinks: durations and lengths are both divided by $\gamma$ .”

What to write

“Time DILATES, $\Delta t=\gamma\Delta t_{proper}$ , whereas lengths CONTRACT, $L=\dfrac{L_{proper}}{\gamma}$ . Since $\gamma\geq 1$ , the measured duration is always LONGER and the measured length always SHORTER.”

0.600c

, the measured duration LENGTHENS to

1.25\,\Delta t_{proper}

and the length SHORTENS to

0.80\,L_{proper}

(green); dividing the duration too (red) is the misreading.

Why: The two effects are opposite because they must compensate each other: the muon sees a contracted distance where the observer on Earth sees a dilated duration, and both conclude that the muon reaches the ground.

Which method to choose

Which relation, according to what the problem describes

The named phenomenon, before any formula

If a wavelength is given and an energy is asked → $E=\dfrac{hc}{\lambda}$ , or the shortcut $\dfrac{1240}{\lambda\text{ in nm}}$ in electron-volts

Example: $400$ nm gives $3.10$ eV
If a metal is illuminated and emits electrons → $E_{photon}=W_{0}+E_{k,max}$ , with $eV_{s}=E_{k,max}$

Example: $3.10-2.30=0.80$ eV, so $V_{s}=0.80$ V
If a hydrogen atom changes level → $E_{photon}=|E_{n_{f}}-E_{n_{i}}|$ with $E_{n}=-\dfrac{13.6}{n^{2}}$ eV

Example: $3\to 2$ gives $1.89$ eV, that is $656$ nm
If a material PARTICLE is associated with a wave → de Broglie, $\lambda=\dfrac{h}{p}$ , with $p=mv$ or $p=\sqrt{2mE_{k}}$

Example: an electron of $100$ eV has a wavelength of $0.12$ nm
If a hot body radiates → Wien for the colour, Stefan for the power, always in kelvins

Example: $\lambda_{max}T=2.898\times 10^{-3}$ m·K
If a nucleus decays or is formed → decay as $\left(\dfrac{1}{2}\right)^{t/T}$ , or mass defect with $1$ u $=931.5$ MeV/ $c^{2}$

Example: helium-4: $7.07$ MeV per nucleon

Two constants are enough to cross the whole chapter in electron-volts: $hc=1240$ eV·nm and $1$ u $=931.5$ MeV/ $c^{2}$ . Memorizing them spares you from converting to joules in almost every exercise.

Radioactive decay: which form to use

What the problem gives and what it asks for

If the duration is a WHOLE multiple of the half-life → powers of two, without a calculator: $\dfrac{1}{2}$ , $\dfrac{1}{4}$ , $\dfrac{1}{8}$

Example: three half-lives leave $12.5\,\%$
If the duration is arbitrary → $N=N_{0}\left(\dfrac{1}{2}\right)^{t/T}$ , or the exponential form with $\lambda=\dfrac{\ln 2}{T}$

Example: the two forms are strictly equivalent
If we look for the elapsed DURATION → isolate with a logarithm: $t=T\dfrac{\ln\left(\frac{N_{0}}{N}\right)}{\ln 2}$

Example: $25\,\%$ of carbon-14 remaining gives $11{,}460$ years
If the problem speaks of ACTIVITY in becquerels → it is proportional to $N$ and decays in exactly the same way

Example: an activity divided by eight corresponds to three half-lives
If the decay constant is asked for → $\lambda=\dfrac{\ln 2}{T}$ , in s $^{-1}$ if $T$ is in seconds

Example: beware of the clash with the wavelength symbol
If a mass or a number of nuclei is asked for → go through the mole: $N=\dfrac{m}{M}\times 6.02\times 10^{23}$

Example: chemistry serves here as the way in

The same letter $\lambda$ stands for the wavelength and for the decay constant: in a problem containing both, it is better to rename one explicitly on the first line.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

A photoelectric problem, written up

When to use it: The problem gives a wavelength and a metal, and asks for the kinetic energy, the stopping potential, the threshold or the speed.

1 Calculate the photon energy in electron-volts with the shortcut $\dfrac{1240}{\lambda}$ , and say so: it is faster and less risky than converting to joules.
2 Compare with the work function: if $E_{photon}<W_{0}$ , conclude immediately that there is no emission and stop there.
3 Otherwise, write $E_{k,max}=E_{photon}-W_{0}$ , then $V_{s}=\dfrac{E_{k,max}}{e}$ , whose numerical value in volts is that of the energy in electron-volts.
4 For a speed, CONVERT to joules first, then $v=\sqrt{\dfrac{2E_{k}}{m}}$ , and check that the result stays well below $c$ .
5 Answer explicitly what the intensity changes: the number of electrons, hence the current, and nothing else.

Concluding sentence

“The $400$ nm photon carries $\dfrac{1240}{400}=3.10$ eV, more than the work function $W_{0}=2.30$ eV: there is emission. The maximum kinetic energy is $0.80$ eV and the stopping potential $0.80$ V. Doubling the intensity would double the current, without changing these two values.”

The trap: Using an energy in electron-volts in $\dfrac{1}{2}mv^{2}$ . The conversion to joules is essential as soon as a mass in kilograms enters the calculation.

Marking: In general 1 point for the photon energy, 1 for the comparison with the threshold, 2 for the kinetic energy and the stopping potential, 2 for the converted speed, 1 for the effect of the intensity.

A dating by radioactive decay

When to use it: The problem gives a remaining proportion, or an activity, and asks for an age.

1 Write the requested ratio in the form $\dfrac{N}{N_{0}}$ , checking that it really is a proportion remaining and not a percentage that has disappeared.
2 If this ratio is an obvious power of two, answer directly in number of half-lives, without a calculator.
3 Otherwise, set $\dfrac{N}{N_{0}}=\left(\dfrac{1}{2}\right)^{t/T}$ and isolate with a logarithm: $t=T\dfrac{\ln\left(\frac{N_{0}}{N}\right)}{\ln 2}$ .
4 Check the order of magnitude: an age shorter than one half-life must correspond to more than half remaining, and conversely.

Concluding sentence

“The sample contains only $25\,\%$ of its initial carbon-14, that is $\left(\dfrac{1}{2}\right)^{2}$ : it has therefore lived exactly two half-lives, that is $2\times 5730=11{,}460$ years.”

The trap: Confusing the fraction remaining with the fraction gone. “ $75\,\%$ of the carbon has decayed” means $25\,\%$ is left, hence two half-lives, and not $0.75$ of a half-life.

Marking: 1 point for the ratio correctly identified, 2 for the equation set up, 1 for the solution, 1 for the check.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The speed compared with that of light

Every particle speed must stay below $3\times 10^{8}$ m/s. An excess almost always signals an electron-volt left in a mechanical formula.

A $v=1.3\times 10^{15}$ m/s: the conversion to joules was forgotten.
The photon energy compared with the visible range

Visible light corresponds to $1.8$ to $3.1$ eV. A photon energy outside this range must correspond to a radiation explicitly named, infrared, ultraviolet or X-rays.

A visible photon reported at $12$ eV: that is far ultraviolet, the wavelength was converted wrongly.
The binding energy per nucleon

It is always between $1$ and $9$ MeV per nucleon, with a maximum around $8.8$ MeV for iron. A result in hundreds of MeV per nucleon is impossible.

A $932$ MeV per nucleon: that is the mass energy of a nucleon, not its binding energy.
The direction of relativistic effects

Check that the duration obtained is LONGER than the proper duration and the length SHORTER than the proper length. Since $\gamma\geq 1$ , there is no other possibility.

A measured duration shorter than the proper duration: $\gamma$ was put in the denominator.
The fraction remaining after n half-lives

Count in powers of two before calculating: $\dfrac{1}{2}$ , $\dfrac{1}{4}$ , $\dfrac{1}{8}$ , $\dfrac{1}{16}$ . A result that does not come close to these for a whole number of half-lives is wrong.

Zero nuclei left after two half-lives: a quarter remains.

The typical problem, taken apart

An illuminated metal: from the photon to the stopping potential

A metal plate with a work function $W_{0}=2.30$ eV is illuminated by violet light of wavelength $\lambda=400$ nm.

Determine the energy of the photons, the maximum kinetic energy of the electrons emitted, the stopping potential and their maximum speed. Then find the threshold wavelength of the metal, and say what doubling the light intensity changes.

The situation: a plate with work function

W_{0}=2.30

eV illuminated at

400

nm, and the microammeter measuring the current of the electrons torn free.

Step 1

Photon energy: $E=\dfrac{1240}{\lambda\text{ (nm)}}=\dfrac{1240}{400}=3.10$ eV.

Why

The shortcut $hc=1240$ eV·nm avoids two conversions and a multiplication of powers of ten. It is the safest way to start, and the result is directly comparable with the work function.

Step 2

Comparison with the threshold: $3.10$ eV $>2.30$ eV, so electrons are indeed emitted.

Why

This comparison is a step in its own right: if it failed, the rest of the problem would have no reason to exist, and that is exactly what the next question on the threshold wavelength tests.

Step 3

Maximum kinetic energy: $E_{k,max}=E_{photon}-W_{0}=3.10-2.30=0.80$ eV. Stopping potential: $V_{s}=\dfrac{E_{k,max}}{e}=0.80$ V.

Why

In electron-volts, the kinetic energy and the stopping potential have the same numerical value, since $E$ in eV divided by the elementary charge gives volts. This is the only place in the course where this coincidence is useful.

Step 4

Maximum speed: convert first, $0.80$ eV $=1.28\times 10^{-19}$ J, then $v=\sqrt{\dfrac{2(1.28\times 10^{-19})}{9.11\times 10^{-31}}}=5.30\times 10^{5}$ m/s.

Why

The conversion is mandatory as soon as a mass in kilograms enters the calculation. The result, about $0.2\,\%$ of the speed of light, confirms in passing that a non-relativistic treatment is legitimate.

Step 5

Threshold wavelength: $W_{0}=\dfrac{1240}{\lambda_{0}}$ gives $\lambda_{0}=\dfrac{1240}{2.30}=539$ nm, a green light.

Why

Any light more RED than green, hence of longer wavelength, will have no effect on this metal, whatever its power. Expressing the threshold as a colour makes the answer checkable at a glance.

Step 6

Corresponding threshold frequency: $f_{0}=\dfrac{W_{0}}{h}=\dfrac{2.30\times 1.6\times 10^{-19}}{6.63\times 10^{-34}}=5.55\times 10^{14}$ Hz. Check: $\dfrac{c}{f_{0}}=540$ nm, which does give back the threshold wavelength.

Why

We recompute the threshold by a different route, through the frequency and through joules, and find the same value: this is the cross-check of the problem.

Step 7

Doubling the intensity: the number of photons doubles, hence the number of electrons emitted and the current double. $E_{k,max}$ , $V_{s}$ and $\lambda_{0}$ stay EXACTLY the same.

Why

This is the answer that sets apart a paper that has understood the chapter: intensity acts on the number, frequency on the energy, and the two never mix.

The conclusion, written out

“The $400$ nm photons carry $3.10$ eV, more than the work function of $2.30$ eV: the electrons leave with at most $0.80$ eV, which gives a stopping potential of $0.80$ V and a maximum speed of $5.30\times 10^{5}$ m/s. The threshold of the metal is at $539$ nm, in the green. Doubling the intensity doubles the current and changes neither the energy of the electrons nor this threshold.”

The classic mistake on this problem: Putting $0.80$ eV directly into $\dfrac{1}{2}mv^{2}$ without conversion. One then obtains a speed of the order of $10^{15}$ m/s, a million times too large and above all greater than that of light.

Learn by heart

• $E=hf=\dfrac{hc}{\lambda}$ , and in electron-volts $E=\dfrac{1240}{\lambda\text{ in nm}}$ .
• Photoelectric: $E_{photon}=W_{0}+E_{k,max}$ and $eV_{s}=E_{k,max}$ ; below the threshold, nothing comes out.
• Intensity changes the NUMBER of electrons; frequency changes their ENERGY.
• Bohr: $E_{n}=-\dfrac{13.6}{n^{2}}$ eV, and the photon carries $|E_{n_{f}}-E_{n_{i}}|$ .
• de Broglie: $\lambda=\dfrac{h}{p}$ , with $p=mv$ or $p=\sqrt{2mE_{k}}$ .
• Wien: $\lambda_{max}T=2.898\times 10^{-3}$ m·K. Stefan: $\dfrac{P}{A}=\sigma T^{4}$ , so $\times 16$ if $T$ doubles.
• After $n$ half-lives, $\left(\dfrac{1}{2}\right)^{n}$ remains, never zero.
• Mass defect: separated nucleons MINUS the nucleus, and $1$ u $=931.5$ MeV/ $c^{2}$ , that is a few MeV per nucleon.
• Compton: $\Delta\lambda=2.43(1-\cos\theta)$ pm, an ABSOLUTE shift, independent of $\lambda$ .
• Relativity: time dilates ( $\times\gamma$ ), lengths contract ( $\div\gamma$ ).

Frequently asked questions

Why doesn't a more intense light tear out faster electrons?

Because an electron absorbs one whole photon, and only one. Increasing the intensity increases the number of photons, hence the number of electrons emitted and the measured current, but each photon keeps the same energy, set by the frequency. The individual energy of the electrons therefore does not change at all.

What happens if the frequency is below the threshold frequency?

Nothing, ever. Each photon carries less energy than the work function of the metal, and since an electron absorbs only one, it cannot escape. Neither a more powerful lamp nor a longer exposure changes anything, which is precisely what classical physics could not explain.

How do I quickly calculate the energy of a photon in electron-volts?

Divide one thousand two hundred and forty by the wavelength in nanometres. This shortcut comes from the product of Planck's constant and the speed of light, and it avoids any conversion. Visible light thus gives photons of one point eight to three point one electron-volts.

Why are the energy levels of the Bohr atom negative?

Because the electron is bound to the nucleus: energy would have to be supplied to detach it. The zero corresponds to the free electron at rest, infinitely far away. It is the gap between two levels, always positive, that gives the energy of the photon emitted or absorbed in a jump.

How much of a sample is left after two half-lives?

A quarter, and not nothing at all. Each half-life halves what is still there, so one half, then one quarter, then one eighth. The decay is exponential and never quite reaches zero, which makes it possible to date objects much older than one half-life.

How do I calculate the binding energy of a nucleus?

Add up the masses of all the protons and all the neutrons taken separately, subtract the mass of the nucleus, and convert this mass defect into energy by multiplying by nine hundred thirty-one point five megaelectron-volts per atomic mass unit. The result is always a few megaelectron-volts per nucleon.

Practise it

Corrected exercises: Modern physics, 203-SN3-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

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Revision sheet: modern physics (203-SN3-RE)

The essentials

The photon and the photoelectric effect

Bohr, de Broglie and the blackbody

The Compton effect: the photon has momentum

The nucleus and relativity

The mistakes that cost marks

1. Expecting faster electrons from more intense light

2. Believing the threshold can be crossed by waiting or by shining harder

3. Mixing electron-volts and joules

4. Subtracting the Bohr levels the wrong way

5. Making the radiated power double when the temperature doubles

6. Believing that two half-lives make everything disappear

7. Calculating a mass defect with the mass of the nucleus alone

8. Expecting a visible Compton effect in visible light

9. Swapping time dilation and length contraction

Which method to choose

Which relation, according to what the problem describes

Radioactive decay: which form to use

How the answer is expected to be written

A photoelectric problem, written up

A dating by radioactive decay

Check before you hand in

The typical problem, taken apart

An illuminated metal: from the photon to the stopping potential

Learn by heart

Frequently asked questions

Why doesn't a more intense light tear out faster electrons?

What happens if the frequency is below the threshold frequency?

How do I quickly calculate the energy of a photon in electron-volts?

Why are the energy levels of the Bohr atom negative?

How much of a sample is left after two half-lives?

How do I calculate the binding energy of a nucleus?

Corrected exercises: Modern physics, 203-SN3-RE

See also

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