Atomic and Nuclear Physics

Atomic Models

Thomson's Plum Pudding Model (1897)

After discovering the electron, Thomson proposed that atoms consist of a uniform positive charge with electrons embedded within it, like plums in a pudding.

Rutherford Scattering (1911)

Rutherford directed alpha particles at a thin gold foil. Most passed through, but some were deflected at large angles, and a few bounced back.

Observations and conclusions:

1. Most alpha particles passed straight through: the atom is mostly empty space.
2. Some were deflected: there is a small, dense, positively charged nucleus.
3. Very few bounced back: the nucleus is very small compared to the atom.

Nuclear atom model: A small, dense, positively charged nucleus surrounded by orbiting electrons.

The Bohr Model (1913)

Bohr proposed a model for the hydrogen atom:

1. Electrons orbit the nucleus in certain allowed orbits (stationary states).
2. Electrons do not radiate energy while in a stationary state.
3. An electron can jump between orbits by absorbing or emitting a photon.
4. The angular momentum is quantised: $mvr = n\dfrac{h}{2\pi}$, where $n = 1, 2, 3, \ldots$

Energy Levels

The energy of an electron in the $n$-th orbit of hydrogen:

$$E_n = -\frac{13.6\mathrm{ eV}}{n^2}$$

Level	$n$	Energy (eV)
Ground state	1	$-13.6$
First excited	2	$-3.40$
Second excited	3	$-1.51$
Third excited	4	$-0.85$
Ionisation	$\infty$	$0$

Photon Emission and Absorption

When an electron transitions from level $n_i$ to level $n_f$:

\Delta E = E_{n_f} - E_{n_i} = hf = \frac`\{hc}`{\lambda}

• Emission ($n_i \gt n_f$): photon is released.
• Absorption ($n_i \lt n_f$): photon is absorbed.

Example

Find the wavelength of light emitted when an electron in hydrogen drops from $n = 3$ to $n = 2$.

$$\Delta E = E_2 - E_3 = -3.40 - (-1.51) = -1.89\mathrm{ eV}$$

(The energy of the photon is $1.89\mathrm{ eV}$.)

\lambda = \frac`\{hc}`{\Delta E} = \frac{1240\mathrm{ eV}\cdot\mathrm{nm}}{1.89\mathrm{ eV}} = 656\mathrm{ nm}

This is the red line in the Balmer series (visible spectrum of hydrogen).

Emission and Absorption Spectra

• Emission spectrum: bright lines on a dark background (photons emitted at specific wavelengths).
• Absorption spectrum: dark lines on a continuous spectrum (photons absorbed at specific wavelengths).
• Each element has a unique spectrum — like a fingerprint.

Spectral Series of Hydrogen

Series	Transitions to	Region
Lyman	$n = 1$	Ultraviolet
Balmer	$n = 2$	Visible
Paschen	$n = 3$	Infrared

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Nuclear Structure

Nucleons

The nucleus contains:

• Protons: positive charge $+e$, mass $\approx 1.673 \times 10^{-27}\mathrm{ kg}$
• Neutrons: no charge, mass $\approx 1.675 \times 10^{-27}\mathrm{ kg}$

The atomic number $Z$ = number of protons (determines the element).

The mass number $A$ = number of protons + number of neutrons.

The neutron number $N = A - Z$.

Nuclear Notation

$${}^A_Z X$$

For example: ${}^{235}_{92}\mathrm{U}$ has 92 protons and 143 neutrons.

Isotopes

Isotopes of an element have the same number of protons but different numbers of neutrons.

Isotope	Protons	Neutrons
${}^{1}_1\mathrm{H}$ (protium)	1	0
${}^{2}_1\mathrm{H}$ (deuterium)	1	1
${}^{3}_1\mathrm{H}$ (tritium)	1	2

Nuclear Forces

• Strong nuclear force: short-range attractive force between nucleons. Overcomes electrostatic repulsion between protons. Range $\approx 1$--$3\mathrm{ fm}$ ($1\mathrm{ fm} = 10^{-15}\mathrm{ m}$).
• Electrostatic (Coulomb) force: repulsive between protons, long range.

Nuclear Radius

The radius of a nucleus is approximately:

$$R = R_0 A^{1/3}$$

where $R_0 \approx 1.2\mathrm{ fm}$.

Nuclear Density

Since $R \propto A^{1/3}$, the volume $V \propto A$. All nuclei have approximately the same density:

$$\rho \approx 2.3 \times 10^{17}\mathrm{ kg/m}^3$$

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Radioactivity

Types of Radiation

Property	Alpha ($\alpha$)	Beta ($\beta^-$)	Gamma ($\gamma$)
Particle	${}^4_2\mathrm{He}$ nucleus	Electron ($e^-$)	Photon
Charge	$+2e$	$-e$	$0$
Mass	$4\mathrm{ u}$	$\approx 0$	$0$
Speed	$\approx 0.05c$	Up to $\approx 0.99c$	$c$
Penetration	Paper	Aluminium (few mm)	Lead (few cm)
Ionising power	High	Medium	Low

Alpha Decay

$${}^A_Z X \to {}^{A-4}_{Z-2} Y + {}^4_2\alpha$$

The daughter nucleus has atomic number reduced by 2 and mass number reduced by 4.

Example

$${}^{238}_{92}\mathrm{U} \to {}^{234}_{90}\mathrm{Th} + {}^4_2\alpha$$

Beta-minus Decay

$${}^A_Z X \to {}^A_{Z+1} Y + {}^0_{-1}\beta + \bar{\nu}_e$$

A neutron converts to a proton, emitting an electron and an antineutrino:

$$n \to p + e^- + \bar{\nu}_e$$

Example

$${}^{14}_6\mathrm{C} \to {}^{14}_7\mathrm{N} + {}^0_{-1}\beta + \bar{\nu}_e$$

Beta-plus Decay (Positron Emission)

$${}^A_Z X \to {}^A_{Z-1} Y + {}^0_{+1}\beta^+ + \nu_e$$

A proton converts to a neutron, emitting a positron and a neutrino:

$$p \to n + e^+ + \nu_e$$

Gamma Decay

The nucleus transitions from a higher energy state to a lower energy state, emitting a gamma photon:

$${}^A_Z X^* \to {}^A_Z X + \gamma$$

No change in $Z$ or $A$.

Half-Life

The half-life $t_{1/2}$ is the time for half of the radioactive nuclei in a sample to decay.

Exponential Decay Law

$$N = N_0 e^{-\lambda t}$$

where:

• $N$ = number of undecayed nuclei at time $t$
• $N_0$ = initial number of nuclei
• $\lambda$ = decay constant

Relationship Between Half-Life and Decay Constant

$$t_{1/2} = \frac{\ln 2}{\lambda}$$

Activity

The activity $A$ is the number of decays per unit time:

A = \lambda N = \frac`\{dN}``\{dt}`

SI unit: becquerel ($\mathrm{Bq}$), where $1\mathrm{ Bq} = 1\mathrm{ decay/s}$.

Example

A sample has a half-life of $5$ days and an initial activity of $800\mathrm{ Bq}$. Find the activity after $15$ days.

$$A = A_0 e^{-\lambda t}$$$$\lambda = \frac{\ln 2}{5} = 0.1386\mathrm{ day}^{-1}$$$$A = 800 \times e^{-0.1386 \times 15} = 800 \times e^{-2.079} = 800 \times 0.125 = 100\mathrm{ Bq}$$

Alternatively: $15$ days $= 3$ half-lives, so $A = 800 \times (1/2)^3 = 100\mathrm{ Bq}$.

Background Radiation

Background radiation comes from natural and artificial sources:

• Cosmic rays
• Radon gas (from rocks and soil)
• Medical procedures (X-rays)
• Nuclear waste and fallout
• Radioactive materials in the Earth

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Nuclear Reactions

Fission

A heavy nucleus splits into two (or more) lighter nuclei, releasing energy and neutrons.

Example:

$${}^{235}_{92}\mathrm{U} + {}^1_0 n \to {}^{141}_{56}\mathrm{Ba} + {}^{92}_{36}\mathrm{Kr} + 3{}^1_0 n + \mathrm{energy}$$

Chain Reaction

The neutrons released in fission can cause further fission events, creating a chain reaction.

• Critical mass: the minimum mass of fissile material needed to sustain a chain reaction.
• Controlled fission: used in nuclear power reactors (control rods absorb neutrons).
• Uncontrolled fission: nuclear weapons.

Fusion

Light nuclei combine to form a heavier nucleus, releasing energy.

Example (proton-proton chain in the Sun):

$$4{}^1_1\mathrm{H} \to {}^4_2\mathrm{He} + 2{}^0_{+1}\beta^+ + 2\nu_e + \mathrm{energy}$$

Conditions for Fusion

• Extremely high temperatures ($\gt 10^7\mathrm{ K}$) to overcome electrostatic repulsion.
• High density to increase collision rate.
• Sufficient confinement time.

Binding Energy per Nucleon

The binding energy per nucleon curve shows:

• Light nuclei (up to Fe-56): fusion increases binding energy per nucleon (releases energy).
• Heavy nuclei (beyond Fe-56): fission increases binding energy per nucleon (releases energy).
• Iron-56 has the highest binding energy per nucleon (most stable nucleus).

Nucleus	Binding Energy per Nucleon (MeV)
${}^2_1\mathrm{H}$	1.11
${}^4_2\mathrm{He}$	7.07
${}^{56}_{26}\mathrm{Fe}$	8.79
${}^{235}_{92}\mathrm{U}$	7.59

Mass Defect and Binding Energy

The mass defect $\Delta m$ is the difference between the mass of a nucleus and the sum of the masses of its individual nucleons:

$$\Delta m = Zm_p + Nm_n - m_{\mathrm{nucleus}}$$

The binding energy is:

$$E_b = \Delta m \cdot c^2$$

Example

Calculate the binding energy of the helium-4 nucleus. ($m_p = 1.00728\mathrm{ u}$, $m_n = 1.00867\mathrm{ u}$, $m_{\alpha} = 4.00151\mathrm{ u}$, $1\mathrm{ u} = 931.5\mathrm{ MeV}/c^2$)

$$\Delta m = 2(1.00728) + 2(1.00867) - 4.00151 = 2.01456 + 2.01734 - 4.00151 = 0.03039\mathrm{ u}$$$$E_b = 0.03039 \times 931.5 = 28.3\mathrm{ MeV}$$

Binding energy per nucleon $= \dfrac{28.3}{4} = 7.08\mathrm{ MeV}$.

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Mass-Energy Equivalence

Einstein's Equation

$$E = mc^2$$

where $c = 3.0 \times 10^8\mathrm{ m/s}$.

This shows that mass and energy are equivalent and can be converted into each other.

Energy-Mass Unit Conversion

$$1\mathrm{ u} = 931.5\mathrm{ MeV}/c^2$$ $$1\mathrm{ eV} = 1.602 \times 10^{-19}\mathrm{ J}$$

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The Photoelectric Effect

Photoelectric Effect

Shine light of different wavelengths and intensities onto a metal surface and observe the emission of photoelectrons. Investigate how frequency, intensity, and the work function affect the maximum kinetic energy.

Phenomenon

When light above a certain frequency shines on a metal surface, electrons are emitted (photoelectrons).

Key Observations

1. Electrons are emitted only if the frequency exceeds a threshold frequency $f_0$.
2. The maximum kinetic energy of photoelectrons depends on frequency, not intensity.
3. Increasing intensity increases the number of photoelectrons (current), not their energy.
4. Emission is instantaneous (no time delay).

Einstein's Explanation

Light consists of photons, each with energy $E = hf$. A single photon can eject one electron if $hf \ge \phi$.

$$E_{k,\max} = hf - \phi$$

where $\phi = hf_0$ is the work function (minimum energy to free an electron from the metal).

Key Equations

$$E_{k,\max} = \frac{1}{2}mv_{\max}^2 = eV_s$$

where $V_s$ is the stopping potential.

$$hf = \phi + E_{k,\max}$$

Graph of $E_{k,\max}$ vs $f$

• Gradient $= h$ (Planck's constant)
• $x$-intercept $= f_0$ (threshold frequency)
• $y$-intercept $= -\phi$

Example

The work function of sodium is $2.28\mathrm{ eV}$. Find the maximum kinetic energy of photoelectrons when light of wavelength $400\mathrm{ nm}$ is incident.

E_{\mathrm{photon}} = \frac`\{hc}`{\lambda} = \frac{1240\mathrm{ eV}\cdot\mathrm{nm}}{400\mathrm{ nm}} = 3.10\mathrm{ eV}$$E_{k,\max} = 3.10 - 2.28 = 0.82\mathrm{ eV}$$

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Wave-Particle Duality

de Broglie Wavelength

All matter has wave-like properties. The de Broglie wavelength of a particle:

\lambda = \frac{h}{p} = \frac{h}`\{mv}`

where $h = 6.626 \times 10^{-34}\mathrm{ J}\cdot\mathrm{s}$ is Planck's constant.

Example

Find the de Broglie wavelength of an electron moving at $2 \times 10^6\mathrm{ m/s}$. ($m_e = 9.109 \times 10^{-31}\mathrm{ kg}$)

$$\lambda = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 2 \times 10^6} = \frac{6.626 \times 10^{-34}}{1.822 \times 10^{-24}} = 3.64 \times 10^{-10}\mathrm{ m} = 0.364\mathrm{ nm}$$

This is comparable to X-ray wavelengths, explaining electron diffraction.

Davisson-Germer Experiment

Confirmed de Broglie's hypothesis by observing the diffraction of electrons by a crystal lattice.

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IB Exam-Style Questions

Question 1 (Paper 1 style)

Uranium-238 undergoes alpha decay. What is the daughter nucleus?

$${}^{238}_{92}\mathrm{U} \to {}^{234}_{90}\mathrm{Th} + {}^4_2\alpha$$

The daughter is thorium-234.

Question 2 (Paper 2 style)

A radioactive isotope has a half-life of 8 hours. A sample initially has an activity of $1200\mathrm{ Bq}$.

(a) Find the activity after 24 hours.

$$n = \frac{24}{8} = 3 \mathrm{ half-lives}$$ $$A = 1200 \times \left(\frac{1}{2}\right)^3 = 150\mathrm{ Bq}$$

(b) Find the decay constant.

$$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{8 \times 3600} = 2.41 \times 10^{-5}\mathrm{ s}^{-1}$$

(c) How many undecayed nuclei remain after 24 hours?

$$A = \lambda N \implies 150 = 2.41 \times 10^{-5} \times N$$ $$N = \frac{150}{2.41 \times 10^{-5}} = 6.22 \times 10^6$$

Question 3 (Paper 2 style)

The work function of a metal surface is $4.5\mathrm{ eV}$.

(a) Find the threshold frequency.

$$f_0 = \frac{\phi}{h} = \frac{4.5 \times 1.602 \times 10^{-19}}{6.626 \times 10^{-34}} = \frac{7.21 \times 10^{-19}}{6.626 \times 10^{-34}} = 1.09 \times 10^{15}\mathrm{ Hz}$$

(b) Find the stopping potential when UV light of frequency $2.0 \times 10^{15}\mathrm{ Hz}$ is incident.

$$E_{k,\max} = hf - \phi = (6.626 \times 10^{-34})(2.0 \times 10^{15}) - 4.5(1.602 \times 10^{-19})$$ $$= 1.325 \times 10^{-18} - 7.21 \times 10^{-19} = 6.04 \times 10^{-20}\mathrm{ J}$$ $$V_s = \frac{E_{k,\max}}{e} = \frac{6.04 \times 10^{-20}}{1.602 \times 10^{-19}} = 0.377\mathrm{ V}$$

Question 4 (Paper 2 style)

Calculate the energy released when ${}^{235}\mathrm{U}$ undergoes fission to produce ${}^{141}\mathrm{Ba}$, ${}^{92}\mathrm{Kr}$, and 3 neutrons.

Masses: ${}^{235}\mathrm{U} = 235.044\mathrm{ u}$, ${}^{141}\mathrm{Ba} = 140.914\mathrm{ u}$, ${}^{92}\mathrm{Kr} = 91.926\mathrm{ u}$, $n = 1.00867\mathrm{ u}$.

$$\Delta m = 235.044 - (140.914 + 91.926 + 3 \times 1.00867)$$ $$= 235.044 - (140.914 + 91.926 + 3.026) = 235.044 - 235.866 = -0.822\mathrm{ u}$$

Wait, that gives negative mass defect. Let me recalculate:

$$\Delta m = 235.044 - 140.914 - 91.926 - 3(1.00867) = 235.044 - 140.914 - 91.926 - 3.026 = -0.822\mathrm{ u}$$

This indicates I should use the neutron as incoming:

$$\Delta m = (235.044 + 1.00867) - (140.914 + 91.926 + 3 \times 1.00867)$$ $$= 236.053 - 235.866 = 0.187\mathrm{ u}$$ $$E = 0.187 \times 931.5 = 174\mathrm{ MeV}$$

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Summary

Concept	Formula
Bohr energy levels	$E_n = \dfrac{-13.6\mathrm{ eV}}{n^2}$
Photon energy	$E = hf = \dfrac{hc}{\lambda}$
Half-life	$t_{1/2} = \dfrac{\ln 2}{\lambda}$
Decay law	$N = N_0 e^{-\lambda t}$
Activity	$A = \lambda N$
Mass-energy	$E = mc^2$
Binding energy	$E_b = \Delta m \cdot c^2$
Photoelectric effect	$E_{k,\max} = hf - \phi$
de Broglie wavelength	$\lambda = \dfrac{h}{mv}$
Nuclear radius	$R = R_0 A^{1/3}$

Exam Strategy

For nuclear physics, always balance your nuclear equations (conservation of $A$ and $Z$). For photoelectric effect problems, check units carefully (eV vs J). For decay calculations, clearly identify the half-life and number of half-lives elapsed.

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Electron Energy Levels and Transitions (Extended)

Energy Level Diagrams

Energy level diagrams show the allowed energies of electrons in an atom:

• The ground state is the lowest energy level ($n = 1$).
• Excited states are higher energy levels ($n = 2, 3, \ldots$).
• The ionisation energy is the energy needed to remove an electron from the ground state to infinity.
• Transition lines represent photon absorption (upward) or emission (downward).

Absorption Spectrum vs Emission Spectrum

Feature	Emission	Absorption
How produced	Hot gas emits light	Cool gas absorbs from continuous source
Appearance	Bright lines on dark background	Dark lines on continuous spectrum
Information	Wavelengths emitted by the element	Wavelengths absorbed by the element
Use	Identifying elements in stars/nebulae	Identifying elements in atmospheres

Series Limits

Each spectral series has a series limit (convergence limit) corresponding to transitions to/from the continuum (ionised state).

For the Lyman series: $\lambda_{\mathrm{limit}} = \dfrac{hc}{13.6\mathrm{ eV}} = 91.2\mathrm{ nm}$.

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Nuclear Physics Extended

Nuclear Binding Energy Curve

The binding energy per nucleon curve reveals:

• Light nuclei (A $\lt 56$): Can release energy through fusion (combining).
• Iron-56: Peak of the curve, most stable nucleus.
• Heavy nuclei (A $\gt 56$): Can release energy through fission (splitting).
• Very light nuclei (A $\lt 4$): Have very low binding energy per nucleon, large energy release in fusion.

Energy from Fission

Example

Calculate the energy released when one nucleus of U-235 fissions into Ba-141 and Kr-92 with 3 neutrons.

Masses: U-235 = $235.0439\mathrm{ u}$, Ba-141 = $140.9139\mathrm{ u}$, Kr-92 = $91.8970\mathrm{ u}$, n = $1.00867\mathrm{ u}$.

Reactants: $235.0439 + 1.00867 = 236.0526\mathrm{ u}$

Products: $140.9139 + 91.8970 + 3(1.00867) = 140.9139 + 91.8970 + 3.0260 = 235.8369\mathrm{ u}$

Mass defect: $236.0526 - 235.8369 = 0.2157\mathrm{ u}$

Energy released: $0.2157 \times 931.5 = 200.9\mathrm{ MeV}$

Nuclear Reactors

Key components:

• Fuel rods: contain fissile material (U-235 or Pu-239).
• Moderator: slows down neutrons (water, heavy water, graphite).
• Control rods: absorb neutrons to control the rate of fission (boron, cadmium).
• Coolant: removes heat from the reactor (water, liquid sodium).
• Shielding: protects workers from radiation (concrete, lead).

Nuclear Waste

Type	Half-life	Handling
High-level (fission products)	Years to centuries	Deep geological disposal
Transuranic waste	Thousands of years	Deep geological disposal
Low-level (contaminated materials)	Days to years	Compaction and shallow burial

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Quantum Mechanics Concepts

Wave-Particle Duality

Both matter and electromagnetic radiation exhibit wave-like and particle-like properties depending on the experiment.

Phenomenon	Wave Nature	Particle Nature
Light	Diffraction, interference	Photoelectric effect, Compton scattering
Electrons	Electron diffraction	Quantised energy levels

Heisenberg Uncertainty Principle

$$\Delta x \cdot \Delta p \ge \frac{h}{2\pi}$$

It is fundamentally impossible to know both the exact position and exact momentum of a particle simultaneously.

Compton Scattering

When X-rays scatter off electrons, the scattered X-ray has a longer wavelength:

$$\Delta\lambda = \frac{h}{m_e c}(1 - \cos\theta)$$

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Additional IB Exam-Style Questions

Question 5 (Paper 2 style)

The isotope Po-210 undergoes alpha decay with a half-life of 138 days.

(a) Write the nuclear equation for the decay.

$${}^{210}_{84}\mathrm{Po} \to {}^{206}_{82}\mathrm{Pb} + {}^4_2\alpha$$

(b) A sample of Po-210 has an initial activity of $800\mathrm{ Bq}$. Find the activity after 1 year (365 days).

Number of half-lives: $n = \dfrac{365}{138} = 2.645$.

$$A = 800 \times \left(\frac{1}{2}\right)^{2.645} = 800 \times 0.162 = 129.6\mathrm{ Bq}$$

(c) Find the number of Po-210 nuclei in the initial sample.

$$\lambda = \frac{\ln 2}{138 \times 86400} = 5.83 \times 10^{-8}\mathrm{ s}^{-1}$$ $$N = \frac{A}{\lambda} = \frac{800}{5.83 \times 10^{-8}} = 1.37 \times 10^{10}$$

(d) Calculate the mass of the initial sample.

$$m = \frac{N \times 210}{N_A} = \frac{1.37 \times 10^{10} \times 210}{6.022 \times 10^{23}} = 4.78 \times 10^{-12}\mathrm{ g}$$

Question 6 (Paper 1 style)

Which of the following radiations has the highest ionising power and the lowest penetrating power?

A. Alpha particles B. Beta particles C. Gamma rays D. X-rays

Answer: A. Alpha particles are the most ionising and the least penetrating.

Question 7 (Paper 2 style)

An electron is accelerated through a potential difference of $500\mathrm{ V}$.

(a) Calculate the kinetic energy of the electron in eV and J.

$$E_k = 500\mathrm{ eV} = 500 \times 1.602 \times 10^{-19} = 8.01 \times 10^{-17}\mathrm{ J}$$

(b) Calculate the de Broglie wavelength of the electron.

$$v = \sqrt{\frac{2E_k}{m_e}} = \sqrt{\frac{2 \times 8.01 \times 10^{-17}}{9.109 \times 10^{-31}}} = \sqrt{1.759 \times 10^{14}} = 1.326 \times 10^7\mathrm{ m/s}$$ \lambda = \frac{h}`\{mv}` = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 1.326 \times 10^7} = \frac{6.626 \times 10^{-34}}{1.208 \times 10^{-23}} = 5.49 \times 10^{-11}\mathrm{ m}

This wavelength is comparable to atomic spacing, suitable for electron diffraction experiments.

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tip

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