3.2 Particles and radiation — printable coverage pack

3.2.1.1 · Constituents of the atom

A proton has charge $+e$ and relative mass $1$ , a neutron has charge $0$ and relative mass $1$ , and an electron has charge $-e$ and a much smaller relative mass of about $1/1836$ .
In nuclide notation ${}^{A}_{Z}X$ , $Z$ is the proton number and $A$ is the nucleon number, so the neutron number is $A-Z$ .
Isotopes have the same proton number but different neutron numbers. An ion has gained or lost electrons; nuclear composition is unchanged by ionisation.
Specific charge is $q/m$ in $\text{C kg}^{-1}$ . Use the net charge and the mass of the whole particle, nucleus or ion, and retain the sign where it matters.

Tier 1 · Easy

1. State the numbers of protons, neutrons and electrons in a ${}^{25}_{12}\text{Mg}^{2+}$ ion.[2 marks]
Answer
- $12$ protons, $13$ neutrons and $10$ electrons
Method: The proton number gives $12$ protons. The neutron number is $25-12=13$ . A $2+$ ion has lost two electrons, so it has $12-2=10$ electrons.

Tier 2 · Standard

1. A helium nucleus has charge $+3.20\times10^{-19}\,\text{C}$ and mass $6.64\times10^{-27}\,\text{kg}$ . Calculate its specific charge.[3 marks]
Answer
- $+4.82\times10^7\,\text{C kg}^{-1}$
Method: Specific charge is $q/m$ . Therefore $q/m=(3.20\times10^{-19})/(6.64\times10^{-27})=4.82\times10^7\,\text{C kg}^{-1}$ . It is positive because the nucleus has positive charge.

Tier 3 · Hard

1. An ion of an isotope with $A=40$ and $Z=20$ has specific charge $+7.19\times10^6\,\text{C kg}^{-1}$ . Use nucleon mass $1.67\times10^{-27}\,\text{kg}$ and $e=1.60\times10^{-19}\,\text{C}$ to determine the ionic charge and the number of electrons in the ion.[5 marks]
Answer
- charge $+3e$ ; $17$ electrons
Method: Approximate the ion mass by its $40$ nucleons: $m=40(1.67\times10^{-27})=6.68\times10^{-26}\,\text{kg}$ . Its charge is $q=(7.19\times10^6)(6.68\times10^{-26})=4.80\times10^{-19}\,\text{C}$ . Dividing by $e$ gives $q/e=(4.80\times10^{-19})/(1.60\times10^{-19})=3$ , so the ion has charge $+3e$ . A neutral atom with $Z=20$ has $20$ electrons; losing three leaves $17$ electrons.

3.2.1.2 · Stable and unstable nuclei

The strong nuclear force is attractive over separations up to about $3\,\text{fm}$ but becomes repulsive below about $0.5\,\text{fm}$ , preventing nucleons from collapsing together.
In alpha decay, nucleon number decreases by $4$ and proton number by $2$ . In $\beta^-$ decay, a neutron becomes a proton, so $A$ is unchanged and $Z$ increases by $1$ .
Write $\beta^-$ decay with both the electron and electron antineutrino: $n\rightarrow p+e^-+\bar{\nu}_e$ . Omitting $\bar{\nu}_e$ is a standard mark-scheme loss.
The neutrino was proposed because a two-body beta decay could not account for the observed continuous electron-energy distribution while conserving energy and momentum.

Tier 1 · Easy

1. Complete the alpha-decay equation ${}^{210}_{84}\text{Po}\rightarrow{}^{206}_{82}\text{Pb}+\,?$ .[1 mark]
Answer
- ${}^{4}_{2}\text{He}$
Method: The missing particle must carry nucleon number $210-206=4$ and proton number $84-82=2$ , so it is an alpha particle, ${}^{4}_{2}\text{He}$ .

Tier 2 · Standard

1. Write the complete equation for the $\beta^-$ decay of ${}^{14}_{6}\text{C}$ and state how the proton number changes.[3 marks]
Answer
- ${}^{14}_{6}\text{C}\rightarrow{}^{14}_{7}\text{N}+e^-+\bar{\nu}_e$ ; proton number increases by $1$
Method: In $\beta^-$ decay a neutron changes into a proton, an electron and an electron antineutrino. The nucleon number therefore stays $14$ , while the proton number increases from $6$ to $7$ : ${}^{14}_{6}\text{C}\rightarrow{}^{14}_{7}\text{N}+e^-+\bar{\nu}_e$ .

Tier 3 · Hard

1. Describe how the strong nuclear force between two nucleons depends on their separation, and explain how this behaviour contributes to a stable nucleus.[5 marks]
Answer
- It is repulsive below about $0.5\,\text{fm}$ , attractive from about $0.5\,\text{fm}$ to $3\,\text{fm}$ , and negligible beyond about $3\,\text{fm}$ ; attraction binds nearby nucleons while short-range repulsion prevents collapse.
Method: Award the marking-point chain: below roughly $0.5\,\text{fm}$ the force is repulsive; from roughly $0.5\,\text{fm}$ to $3\,\text{fm}$ it is attractive; beyond roughly $3\,\text{fm}$ it is negligible. The attractive region binds neighbouring nucleons and can overcome proton-proton electrostatic repulsion at nuclear separations. The very-short-range repulsion prevents nucleons from collapsing into the same position, giving a stable separation.

3.2.1.3 · Particles, antiparticles and photons

Every particle has an antiparticle with the same mass and rest energy but opposite charge and opposite additive quantum numbers; a neutral antiparticle is not necessarily identical to its particle.
A photon has energy $E=hf=hc/\lambda$ , where $h=6.63\times10^{-34}\,\text{J s}$ and $c=3.00\times10^8\,\text{m s}^{-1}$ .
Annihilation converts a particle and antiparticle into photons; pair production is the reverse and needs at least the combined rest energy of the pair plus a nearby body to conserve momentum.
For a slow electron-positron pair annihilating into two photons, each photon has approximately $0.511\,\text{MeV}$ . Do not assign the full $1.022\,\text{MeV}$ to each photon.

Tier 1 · Easy

1. Calculate the energy of a photon of frequency $6.0\times10^{14}\,\text{Hz}$ . Use $h=6.63\times10^{-34}\,\text{J s}$ .[2 marks]
Answer
- $4.0\times10^{-19}\,\text{J}$
Method: Use $E=hf$ : $E=(6.63\times10^{-34})(6.0\times10^{14})=3.978\times10^{-19}\,\text{J}$ , which is $4.0\times10^{-19}\,\text{J}$ to two significant figures.

Tier 2 · Standard

1. A slow electron and a slow positron annihilate to produce two identical photons. Each electron has rest energy $0.511\,\text{MeV}$ . Calculate the wavelength of either photon. Use $h=6.63\times10^{-34}\,\text{J s}$ , $c=3.00\times10^8\,\text{m s}^{-1}$ and $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .[3 marks]
Answer
- $2.43\times10^{-12}\,\text{m}$
Method: Two photons share the two rest energies, so each photon has $0.511\,\text{MeV}$ . In joules this is $E=0.511\times10^6\times1.60\times10^{-19}=8.176\times10^{-14}\,\text{J}$ . Hence $\lambda=hc/E=(6.63\times10^{-34})(3.00\times10^8)/(8.176\times10^{-14})=2.43\times10^{-12}\,\text{m}$ .

Tier 3 · Hard

1. A photon of wavelength $3.00\times10^{-13}\,\text{m}$ produces an electron-positron pair near a nucleus. Neglecting the nucleus's recoil energy, calculate the total kinetic energy of the pair in $\text{MeV}$ and explain why the nucleus is required. Use $h=6.63\times10^{-34}\,\text{J s}$ , $c=3.00\times10^8\,\text{m s}^{-1}$ , $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ and electron rest energy $0.511\,\text{MeV}$ .[5 marks]
Answer
- $3.12\,\text{MeV}$ ; the nucleus recoils to conserve momentum
Method: The photon energy is $E=hc/\lambda=(6.63\times10^{-34})(3.00\times10^8)/(3.00\times10^{-13})=6.63\times10^{-13}\,\text{J}$ . This is $(6.63\times10^{-13})/(1.60\times10^{-13})=4.14\,\text{MeV}$ . Creating the two rest masses requires $2(0.511)=1.022\,\text{MeV}$ , leaving $4.14-1.022=3.12\,\text{MeV}$ as total kinetic energy. A single photon has momentum, so the nearby nucleus must recoil and take momentum for momentum to be conserved.

3.2.1.4 · Particle interactions

The four fundamental interactions are gravitational, electromagnetic, weak and strong. Particle reactions here are described through exchange particles rather than action at a distance.
The electromagnetic interaction uses virtual photons. In this specification the weak interaction covers $\beta^-$ decay, $\beta^+$ decay, electron capture and electron-proton collisions, using $W^+$ or $W^-$ bosons.
At each interaction vertex, charge must be conserved. For $\beta^-$ decay the sequence is $n\rightarrow p+W^-$ followed by $W^-\rightarrow e^-+\bar{\nu}_e$ .
An exchange-particle diagram must show the correct incoming and outgoing particles, the exchanged particle and its direction; naming only the force is not a complete interaction description.

Tier 1 · Easy

1. State the fundamental interaction and exchange particle involved when a neutron undergoes $\beta^-$ decay.[2 marks]
Answer
- weak interaction; $W^-$ boson
Method: Beta decay is caused by the weak interaction. In $\beta^-$ decay the neutron emits a negatively charged exchange boson, so the exchange particle is $W^-$ .

Tier 2 · Standard

1. Complete the weak-interaction equation $e^-+p\rightarrow n+\,?$ and identify the exchange particle involved.[3 marks]
Answer
- $e^-+p\rightarrow n+\nu_e$ ; a $W^-$ boson
Method: Initial charge is $-1+1=0$ , so the missing particle is neutral. Electron lepton number starts at $+1$ , so the outgoing neutral lepton must be $\nu_e$ , also with electron lepton number $+1$ . The electron emits $W^-$ as it becomes $\nu_e$ , and the proton absorbs $W^-$ to become a neutron. Thus $e^-+p\rightarrow n+\nu_e$ is mediated by $W^-$ .

Tier 3 · Hard

1. Describe $\beta^+$ decay as two interaction vertices involving an exchange particle, and use charge at each vertex to justify the sign of that exchange particle.[5 marks]
Answer
- $p\rightarrow n+W^+$ followed by $W^+\rightarrow e^++\nu_e$ ; charge is conserved at both vertices
Method: At the first vertex a proton of charge $+1$ becomes a neutron of charge $0$ , so the emitted exchange particle must carry charge $+1$ : $p\rightarrow n+W^+$ . At the second vertex the $W^+$ produces a positron of charge $+1$ and a neutral electron neutrino: $W^+\rightarrow e^++\nu_e$ . Charge is therefore $+1=0+(+1)$ at the first vertex and $+1=(+1)+0$ at the second; the complete reaction is $p\rightarrow n+e^++\nu_e$ .

3.2.1.5 · Classification of particles

Hadrons experience the strong interaction. Baryons and antibaryons contain three quarks or three antiquarks and have baryon number $+1$ or $-1$ ; mesons contain a quark-antiquark pair and have baryon number $0$ .
Required baryons are the proton and neutron, and required mesons are pions and kaons. The proton is the only stable baryon; the pion acts as the exchange particle of the strong nuclear force.
Leptons do not experience the strong interaction. Required leptons are the electron, muon, electron neutrino and muon neutrino, together with their antiparticles.
Strange particles are produced through the strong interaction, usually with total strangeness conserved, but decay through the weak interaction, where strangeness may change by $0$ , $+1$ or $-1$ .
Electron-family and muon-family lepton numbers are conserved separately. A common error is to check only a single total lepton number.

Tier 1 · Easy

1. Classify the proton, pion and muon.[2 marks]
Answer
- proton: baryon; pion: meson; muon: lepton
Method: The proton is a three-quark hadron, so it is a baryon. The pion is a quark-antiquark hadron, so it is a meson. The muon is not a hadron and belongs to the lepton family.

Tier 2 · Standard

1. Complete the decay $\mu^-\rightarrow e^-+\,?+\,?$ using an electron-type neutrino and a muon-type neutrino, and state the particle class shared by all four particles.[3 marks]
Answer
- $\mu^-\rightarrow e^-+\bar{\nu}_e+\nu_\mu$ ; all are leptons
Method: Initially $L_\mu=+1$ and $L_e=0$ . The outgoing electron supplies $L_e=+1$ , so an electron antineutrino with $L_e=-1$ is also needed. A muon neutrino supplies the initial $L_\mu=+1$ . Thus $\mu^-\rightarrow e^-+\bar{\nu}_e+\nu_\mu$ , and every particle shown is a lepton or antilepton.

Tier 3 · Hard

1. In the strong interaction $p+p\rightarrow p+\Lambda^0+K^+$ , the supplied data are $B(\Lambda^0)=1$ , $S(\Lambda^0)=-1$ , $B(K^+)=0$ and $S(K^+)=+1$ . Explain how the products illustrate the classification and paired production of strange particles.[5 marks]
Answer
- $\Lambda^0$ is a strange baryon and $K^+$ a strange meson; total $B$ remains $2$ , total charge remains $+2$ , and the produced strangeness sums to $0$ .
Method: The initial baryon number is $1+1=2$ . The final value is $1+1+0=2$ , so $\Lambda^0$ is classified as a baryon while $K^+$ is a meson. Charge is also conserved: $+2=(+1)+0+(+1)$ . Initial strangeness is $0$ , and the new particles have $-1$ and $+1$ , giving total strangeness $0$ . Their opposite strangeness therefore demonstrates paired production in a strong interaction.

3.2.1.6 · Quarks and antiquarks

Up, down and strange quarks have charges $+\frac{2}{3}e$ , $-\frac{1}{3}e$ and $-\frac{1}{3}e$ respectively; antiquarks have opposite charge and quantum numbers.
Each quark has baryon number $+\frac{1}{3}$ and each antiquark $-\frac{1}{3}$ . A strange quark has strangeness $-1$ and an antistrange quark has strangeness $+1$ .
Know $p=uud$ and $n=udd$ ; the corresponding antibaryons contain $\bar{u}\bar{u}\bar{d}$ and $\bar{u}\bar{d}\bar{d}$ .
Mesons are quark-antiquark pairs. Required examples include $\pi^+=u\bar{d}$ , $\pi^-=d\bar{u}$ , $K^+=u\bar{s}$ and $K^0=d\bar{s}$ , with antiparticles obtained by replacing every quark with its antiquark.

Tier 1 · Easy

1. State the quark composition of a proton and show that it has charge $+e$ .[2 marks]
Answer
- $uud$ ; $+\frac{2}{3}e+\frac{2}{3}e-\frac{1}{3}e=+e$
Method: A proton is $uud$ . Adding the quark charges gives $+\frac{2}{3}e+\frac{2}{3}e-\frac{1}{3}e=+e$ , as required.

Tier 2 · Standard

1. A meson has quark composition $u\bar{s}$ . Determine its charge, baryon number and strangeness, and identify the meson.[3 marks]
Answer
- charge $+e$ , baryon number $0$ , strangeness $+1$ ; $K^+$
Method: The charge is $+\frac{2}{3}e+\frac{1}{3}e=+e$ . Its baryon number is $+\frac{1}{3}-\frac{1}{3}=0$ . The antistrange quark gives strangeness $+1$ . A meson with composition $u\bar{s}$ is therefore $K^+$ .

Tier 3 · Hard

1. Describe neutron $\beta^-$ decay in terms of a quark change and a $W$ boson. Verify charge conservation at both vertices.[5 marks]
Answer
- $d\rightarrow u+W^-$ , changing $udd$ to $uud$ , followed by $W^-\rightarrow e^-+\bar{\nu}_e$
Method: A neutron is $udd$ and a proton is $uud$ , so one down quark changes into an up quark: $d\rightarrow u+W^-$ . Charge at this vertex is $-\frac{1}{3}e=+\frac{2}{3}e-e$ . The remaining quarks are spectators, so $udd$ becomes $uud$ . The second vertex is $W^-\rightarrow e^-+\bar{\nu}_e$ , with charge $-e=-e+0$ . The full decay is therefore $n\rightarrow p+e^-+\bar{\nu}_e$ .

3.2.1.7 · Applications of conservation laws

Charge, baryon number, each lepton-family number, energy and momentum are conserved in every particle interaction.
Strangeness is conserved in strong interactions but may change by $0$ , $+1$ or $-1$ in a weak interaction. Use supplied particle data when an unfamiliar particle appears.
In $\beta^-$ decay a down quark changes to an up quark; in $\beta^+$ decay an up quark changes to a down quark.
A reaction passing charge conservation alone is not necessarily possible. Tabulate initial and final charge, baryon number, electron lepton number, muon lepton number and, where relevant, strangeness.

Tier 1 · Easy

1. Show that charge, baryon number and lepton number are conserved in $p+p\rightarrow p+n+\pi^+$ .[2 marks]
Answer
- charge: $+2e=+2e$ ; baryon number: $2=2$ ; lepton number: $0=0$
Method: Initially the charge is $+e+e=+2e$ ; finally it is $+e+0+e=+2e$ . Initially $B=1+1=2$ ; finally $B=1+1+0=2$ . No leptons occur on either side, so lepton number is $0$ before and after.

Tier 2 · Standard

1. Use conservation of charge, baryon number and electron lepton number to determine $X$ in $n\rightarrow p+e^-+X$ .[4 marks]
Answer
- $X=\bar{\nu}_e$
Method: Charge is conserved because $0=(+1)+(-1)+0$ , so $X$ is neutral. Baryon number is $1=1+0+0$ , so $X$ is not a baryon. Electron lepton number initially is $0$ but the electron contributes $+1$ , so $X$ must contribute $-1$ . The neutral particle with $L_e=-1$ is the electron antineutrino, $X=\bar{\nu}_e$ .

Tier 3 · Hard

1. Two proposed reactions are $\nu_e+n\rightarrow p+e^-$ and $\bar{\nu}_e+n\rightarrow p+e^-$ . Determine which can occur by checking charge, baryon number, electron lepton number, energy and momentum.[6 marks]
Answer
- $\nu_e+n\rightarrow p+e^-$ can occur if energy and momentum permit; the antineutrino reaction violates electron lepton-number conservation
Method: For both proposals, charge is $0+0=+1-1=0$ and baryon number is $0+1=1+0=1$ . For $\nu_e+n\rightarrow p+e^-$ , electron lepton number is $+1+0=0+(+1)$ , so it is conserved. For $\bar{\nu}_e+n\rightarrow p+e^-$ , it would be $-1+0$ initially but $0+(+1)$ finally, so it changes from $-1$ to $+1$ and the reaction is forbidden. The first reaction is allowed by these quantum-number tests, but it can occur only when the initial energy is sufficient and total momentum can also be conserved.

3.2.2.1 · The photoelectric effect

One photon transfers all its energy $hf$ to one surface electron. Emission occurs only when $hf\geq\phi$ , so the threshold frequency is $f_0=\phi/h$ .
The maximum photoelectron kinetic energy is $E_{\text{k,max}}=hf-\phi$ . The stopping potential satisfies $eV_s=E_{\text{k,max}}$ .
Increasing frequency above threshold increases maximum kinetic energy and stopping potential. Increasing intensity at fixed frequency increases the emission rate but not the maximum kinetic energy.
Convert work function consistently between electronvolts and joules using $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ . A negative value from $hf-\phi$ means no photoelectrons are emitted.

Tier 1 · Easy

1. A metal has work function $2.4\,\text{eV}$ . Calculate its threshold frequency. Use $h=6.63\times10^{-34}\,\text{J s}$ and $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .[2 marks]
Answer
- $5.8\times10^{14}\,\text{Hz}$
Method: Convert the work function: $\phi=2.4\times1.60\times10^{-19}=3.84\times10^{-19}\,\text{J}$ . At threshold $hf_0=\phi$ , so $f_0=(3.84\times10^{-19})/(6.63\times10^{-34})=5.79\times10^{14}\,\text{Hz}$ , or $5.8\times10^{14}\,\text{Hz}$ .

Tier 2 · Standard

1. Radiation of frequency $8.2\times10^{14}\,\text{Hz}$ illuminates a metal of work function $2.1\,\text{eV}$ . Calculate the maximum kinetic energy in $\text{eV}$ and the stopping potential. Use $h=6.63\times10^{-34}\,\text{J s}$ and $e=1.60\times10^{-19}\,\text{C}$ .[3 marks]
Answer
- $1.3\,\text{eV}$ ; $1.3\,\text{V}$
Method: The photon energy in electronvolts is $hf/e=(6.63\times10^{-34})(8.2\times10^{14})/(1.60\times10^{-19})=3.40\,\text{eV}$ . Hence $E_{\text{k,max}}=3.40-2.1=1.30\,\text{eV}$ . Since one electronvolt corresponds to an electron crossing one volt, $eV_s=E_{\text{k,max}}$ gives $V_s=1.30\,\text{V}$ .

Tier 3 · Hard

1. Light of wavelength $420\,\text{nm}$ illuminates a metal with work function $2.30\,\text{eV}$ . Calculate the stopping potential and state the effect on emission rate and stopping potential when the light intensity is doubled at the same wavelength. Use $h=6.63\times10^{-34}\,\text{J s}$ , $c=3.00\times10^8\,\text{m s}^{-1}$ and $e=1.60\times10^{-19}\,\text{C}$ .[5 marks]
Answer
- $0.66\,\text{V}$ ; emission rate doubles and stopping potential is unchanged
Method: The photon energy is $E=hc/\lambda=(6.63\times10^{-34})(3.00\times10^8)/(420\times10^{-9})=4.74\times10^{-19}\,\text{J}=2.96\,\text{eV}$ . Thus $E_{\text{k,max}}=2.96-2.30=0.66\,\text{eV}$ and $V_s=0.66\,\text{V}$ . Doubling intensity doubles the photon arrival rate, so the photoelectron emission rate doubles. Each photon still has the same energy, so maximum kinetic energy and stopping potential are unchanged.

3.2.2.2 · Collisions of electrons with atoms

Excitation raises an atomic electron to a higher bound energy level; ionisation removes it completely. The incident electron must transfer at least the relevant discrete excitation or ionisation energy.
After an inelastic collision, energy not transferred to the atom remains as kinetic energy of the incident electron; transfers smaller than an allowed energy gap cannot excite the atom.
An electron accelerated through potential difference $V$ gains energy $eV$ , equal to $V\,\text{eV}$ when expressed in electronvolts.
In a fluorescent tube, accelerated electrons excite mercury atoms; de-excitation produces ultraviolet photons, which excite the coating so that it emits visible photons.

Tier 1 · Easy

1. Convert an electron energy of $18\,\text{eV}$ into joules. Use $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .[1 mark]
Answer
- $2.9\times10^{-18}\,\text{J}$
Method: $18\,\text{eV}=18(1.60\times10^{-19})=2.88\times10^{-18}\,\text{J}$ , which is $2.9\times10^{-18}\,\text{J}$ to two significant figures.

Tier 2 · Standard

1. An electron with kinetic energy $12.0\,\text{eV}$ collides with an atom in its ground state. Excited states are $4.0\,\text{eV}$ and $10.5\,\text{eV}$ above the ground state, and the ionisation energy is $13.6\,\text{eV}$ . Determine the greatest possible excitation and the electron kinetic energy immediately afterwards.[3 marks]
Answer
- $10.5\,\text{eV}$ excitation; $1.5\,\text{eV}$ remaining
Method: The electron cannot ionise the atom because $12.0<13.6\,\text{eV}$ . It can supply the larger available discrete excitation energy, $10.5\,\text{eV}$ . Conservation of energy leaves $12.0-10.5=1.5\,\text{eV}$ as the incident electron's kinetic energy.

Tier 3 · Hard

1. Mercury atoms in a fluorescent tube have an excitation energy of $7.7\,\text{eV}$ . Determine the minimum accelerating potential needed for an electron to cause this excitation, then explain how the collision ultimately produces visible light from the tube coating.[5 marks]
Answer
- $7.7\,\text{V}$ ; electron impact excites mercury, mercury emits ultraviolet photons on de-excitation, and the coating absorbs these and emits visible photons
Method: An electron accelerated through $V$ volts gains $V\,\text{eV}$ , so the minimum potential for a $7.7\,\text{eV}$ excitation is $7.7\,\text{V}$ . In an inelastic collision the electron transfers this discrete energy to a mercury atom. When the atom returns to a lower level it emits an ultraviolet photon. The fluorescent coating absorbs that ultraviolet photon, becomes excited, and emits lower-energy visible photons as it de-excites.

3.2.2.3 · Energy levels and photon emission

Atomic electrons occupy discrete energy levels. A line emission spectrum is evidence that only particular downward transitions, and therefore particular photon energies, are allowed.
For a downward transition from $E_1$ to $E_2$ , the emitted photon satisfies $hf=E_1-E_2$ with $E_1>E_2$ ; use the positive magnitude of the energy difference.
An upward transition requires absorption of a photon whose energy exactly equals the level gap. A photon with an intermediate energy is not partly absorbed.
When levels are quoted in electronvolts, find the gap in electronvolts and convert it to joules before using $f=\Delta E/h$ or $\lambda=hc/\Delta E$ .

Tier 1 · Easy

1. An electron falls from an energy level at $-1.5\,\text{eV}$ to one at $-3.4\,\text{eV}$ . State the photon energy.[1 mark]
Answer
- $1.9\,\text{eV}$
Method: The emitted energy is the positive level difference: $\Delta E=(-1.5)-(-3.4)=1.9\,\text{eV}$ .

Tier 2 · Standard

1. Calculate the wavelength emitted when an electron falls from $-1.51\,\text{eV}$ to $-3.40\,\text{eV}$ . Use $h=6.63\times10^{-34}\,\text{J s}$ , $c=3.00\times10^8\,\text{m s}^{-1}$ and $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .[3 marks]
Answer
- $6.58\times10^{-7}\,\text{m}$
Method: The energy gap is $(-1.51)-(-3.40)=1.89\,\text{eV}=1.89(1.60\times10^{-19})=3.024\times10^{-19}\,\text{J}$ . Therefore $\lambda=hc/\Delta E=(6.63\times10^{-34})(3.00\times10^8)/(3.024\times10^{-19})=6.58\times10^{-7}\,\text{m}$ .

Tier 3 · Hard

1. An atom has energy levels $0$ , $-1.2$ , $-3.0$ and $-5.0\,\text{eV}$ . Electrons are raised to the $0\,\text{eV}$ level and can return by any sequence. Determine the number of distinct emission lines and calculate the longest wavelength. Use $h=6.63\times10^{-34}\,\text{J s}$ , $c=3.00\times10^8\,\text{m s}^{-1}$ and $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .[5 marks]
Answer
- $6$ lines; $1.04\times10^{-6}\,\text{m}$
Method: Four levels give one line for each pair of levels, so the distinct gaps are $1.2$ , $3.0$ , $5.0$ , $1.8$ , $3.8$ and $2.0\,\text{eV}$ : six lines. The longest wavelength comes from the smallest gap, $1.2\,\text{eV}=1.92\times10^{-19}\,\text{J}$ . Thus $\lambda_{\max}=hc/\Delta E=(6.63\times10^{-34})(3.00\times10^8)/(1.92\times10^{-19})=1.04\times10^{-6}\,\text{m}$ .

3.2.2.4 · Wave-particle duality

Electron diffraction is evidence that matter particles have wave properties, while the photoelectric effect is evidence that electromagnetic radiation transfers energy in particle-like photons.
The de Broglie wavelength is $\lambda=h/p=h/(mv)$ for a non-relativistic particle. Use momentum in $\text{kg m s}^{-1}$ to obtain wavelength in metres.
Greater particle momentum gives a shorter de Broglie wavelength and therefore less diffraction for the same aperture or crystal spacing.
For an electron accelerated from rest through potential $V$ , $eV=\frac{1}{2}mv^2$ and hence $\lambda=h/\sqrt{2meV}$ . Do not substitute the voltage itself as an energy in joules.

Tier 1 · Easy

1. Calculate the de Broglie wavelength of a particle with momentum $3.0\times10^{-24}\,\text{kg m s}^{-1}$ . Use $h=6.63\times10^{-34}\,\text{J s}$ .[2 marks]
Answer
- $2.2\times10^{-10}\,\text{m}$
Method: Use $\lambda=h/p$ : $\lambda=(6.63\times10^{-34})/(3.0\times10^{-24})=2.21\times10^{-10}\,\text{m}$ , or $2.2\times10^{-10}\,\text{m}$ .

Tier 2 · Standard

1. Electrons travel at $2.5\times10^6\,\text{m s}^{-1}$ . Calculate their de Broglie wavelength and state how doubling their speed affects the wavelength. Use electron mass $9.11\times10^{-31}\,\text{kg}$ and $h=6.63\times10^{-34}\,\text{J s}$ .[3 marks]
Answer
- $2.9\times10^{-10}\,\text{m}$ ; doubling speed halves the wavelength
Method: The momentum is $p=mv=(9.11\times10^{-31})(2.5\times10^6)=2.28\times10^{-24}\,\text{kg m s}^{-1}$ . Hence $\lambda=h/p=(6.63\times10^{-34})/(2.28\times10^{-24})=2.91\times10^{-10}\,\text{m}$ . Since $\lambda=h/(mv)$ , doubling $v$ doubles momentum and halves the wavelength.

Tier 3 · Hard

1. Electrons are accelerated from rest through $150\,\text{V}$ and then diffract from a crystal. Derive an expression for their de Broglie wavelength in terms of $V$ , calculate it, and explain how increasing $V$ changes the diffraction. Use $e=1.60\times10^{-19}\,\text{C}$ , electron mass $9.11\times10^{-31}\,\text{kg}$ and $h=6.63\times10^{-34}\,\text{J s}$ .[5 marks]
Answer
- $\lambda=\dfrac{h}{\sqrt{2meV}}=1.00\times10^{-10}\,\text{m}$ ; increasing $V$ reduces the wavelength and the diffraction angle
Method: Energy conservation gives $eV=\frac{1}{2}mv^2$ . Since $p=mv$ , squaring gives $p^2=m^2v^2=2meV$ , so $p=\sqrt{2meV}$ and $\lambda=h/\sqrt{2meV}$ . Substitution gives $\lambda=(6.63\times10^{-34})/\sqrt{2(9.11\times10^{-31})(1.60\times10^{-19})(150)}=1.00\times10^{-10}\,\text{m}$ . Increasing $V$ increases momentum as $\sqrt{V}$ , so $\lambda$ decreases as $1/\sqrt{V}$ and the diffraction angle or ring diameter decreases.