Exercises — De Broglie hypothesis — matter waves λ = h - p

WHAT we do: use the master form directly — nothing to convert, momentum is already given. $\lambda =\frac{h}{p}=\frac{6.626\times 10^{-34}}{3.3\times 10^{-24}}.$ WHY this form: the problem hands us $p$, so $\lambda =h/p$ needs no rewriting. $\lambda =2.008\times 10^{-10}\text{m}\approx 2.0\text{A˚}.$ That is about one atom wide — exactly why electrons diffract off crystals.

WHAT we do: recognise the "accelerated electron" situation ⇒ use the ready-made shortcut. WHY this form: for an electron through voltage $V$, the constants are pre-baked into $12.27/\sqrt{V}$. $\lambda =\frac{12.27}{\sqrt{100}}=\frac{12.27}{10}=1.227\text{A˚}.$

WHAT we do: a photon is massless, so $\lambda =h/(mv)$ is forbidden (it would divide by zero). Use $p=E/c$. WHY: the photon carries momentum through $p=E/c$, giving $\lambda =hc/E$. $\lambda =\frac{hc}{E}=\frac{(6.626\times 10^{-34})(3\times 10^8)}{3.0\times 10^{-19}}=6.626\times 10^{-7}\text{m}\approx 663\text{nm}.$ Visible red light — consistent with a real photon.

WHAT: we are told $K$, not $v$. Convert energy to momentum with $p=\sqrt{2mK}$. WHY $p=\sqrt{2mK}$: because $K=\frac{1}{2}m{v}^2=\frac{p^2}{2m}$, so $p^2=2mK$. $p=\sqrt{2(1.673\times 10^{-27})(1.6\times 10^{-16})}=\sqrt{5.35\times 10^{-43}}=7.32\times 10^{-22}\text{kg⋅m/s}.$ $\lambda =\frac{6.626\times 10^{-34}}{7.32\times 10^{-22}}=9.05\times 10^{-13}\text{m}\approx 9.05\times 10^{-3}\text{A˚}.$ (Since $1\text{A˚}=10^{-10}\text{m}$, we divide $9.05\times 10^{-13}$ by $10^{-10}$ to get ångström.)

WHAT: speed and mass given, slow object ⇒ $p=mv$. WHY $p=mv$: the grain moves far below light-speed, so momentum is just mass times velocity — no relativistic correction needed. $p=(1.0\times 10^{-9})(1.0\times 10^{-3})=1.0\times 10^{-12}\text{kg⋅m/s}.$ $\lambda =\frac{6.626\times 10^{-34}}{1.0\times 10^{-12}}=6.626\times 10^{-22}\text{m}.$ Comment: far, far smaller than any atom ($10^{-22}$ vs $10^{-10}$ m). No wave behaviour is ever observable — the grain is "classical."

WHAT: charge through voltage ⇒ $K=qV$, then $\lambda =h/\sqrt{2mqV}$. We cannot use the $12.27/\sqrt{V}$ shortcut — that was electron-only (specific $m$ and $q$). WHY not the shortcut: $12.27/\sqrt{V}$ hides $m_e$ and $e$ inside it; this particle has different mass and charge. $2mqV=2(6.64\times 10^{-27})(3.2\times 10^{-19})(1000)=4.25\times 10^{-42}.$ $p=\sqrt{4.25\times 10^{-42}}=2.06\times 10^{-21}\text{kg⋅m/s},\lambda =\frac{6.626\times 10^{-34}}{2.06\times 10^{-21}}=3.21\times 10^{-13}\text{m}.$

WHAT: at fixed $v$, $\lambda =\frac{h}{mv}\propto \frac{1}{m}$. WHY ratio not numbers: the $h$ and $v$ cancel, leaving pure mass dependence. $\frac{{\lambda }_e}{{\lambda }_p}=\frac{m_p}{m_e}=\frac{1.673\times 10^{-27}}{9.11\times 10^{-31}}\approx 1836.$ The electron is wavier, by a factor $\approx 1836$.

WHAT: at fixed $K$, $\lambda =\frac{h}{\sqrt{2mK}}\propto \frac{1}{\sqrt{m}}$. WHY the change: energy is shared differently than speed — the mass now sits under a square root. $\frac{{\lambda }_e}{{\lambda }_p}=\sqrt{\frac{m_p}{m_e}}=\sqrt{1836}\approx \mathrm{42.8.}$ The electron is still wavier, but only by $\approx 42.8$, not $1836$.

WHAT: $\lambda \propto 1/\sqrt{V}$, so halving $\lambda$ needs $\sqrt{V}$ to double, i.e. $V$ to quadruple. WHY: $\lambda =12.27/\sqrt{V}\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\sqrt{\frac{V_1}{V_2}}$. $\frac{0.5}{1.0}=\sqrt{\frac{V_1}{V_2}}\Rightarrow \frac{V_1}{V_2}=0.25\Rightarrow V_2=4V_1.$ Concretely $V_1=(12.27/1.0{)}^2\approx 150.6\text{V}$, so $V_2\approx 602\text{V}$.

WHAT: the standing-wave condition says the circumference must hold exactly $n$ ripples. For $n=1$: $\lambda =2\pi r=2\pi (0.529\times 10^{-10})=3.32\times 10^{-10}\text{m}=3.32\text{A˚}.$ WHY it looks like this (see figure below): a wave that does not fit a whole number of times would overlap itself out of phase and cancel — only whole-number fits survive, which is exactly Bohr's quantisation reborn as wave interference. Cross-check via momentum, using the reduced constant $\hslash =h/2\pi$ defined above: Bohr's rule $mvr=n\hslash$ gives $p=mv=\frac{n\hslash }{r}=\frac{1.055\times 10^{-34}}{0.529\times 10^{-10}}=1.994\times 10^{-24}$, so $\lambda =h/p=\frac{6.626\times 10^{-34}}{1.994\times 10^{-24}}=3.32\text{A˚}$ — same answer. ✓

WHAT: temperature ⇒ energy through $K=\frac{3}{2}{k}_{B}T$, then $\lambda =h/\sqrt{2mK}$. WHY $\frac{3}{2}{k}_{B}T$: at temperature $T$ a free particle carries, on average, $\frac{1}{2}{k}_{B}T$ of kinetic energy per direction of motion, and there are three directions — so $\frac{3}{2}{k}_{B}T$ total. $K=\frac{3}{2}(1.38\times 10^{-23})(300)=6.21\times 10^{-21}\text{J}.$ $p=\sqrt{2(1.675\times 10^{-27})(6.21\times 10^{-21})}=\sqrt{2.08\times 10^{-47}}=4.56\times 10^{-24}.$ $\lambda =\frac{6.626\times 10^{-34}}{4.56\times 10^{-24}}=1.45\times 10^{-10}\text{m}\approx 1.45\text{A˚}.$ Why it matters: $\approx 1.4\text{A˚}$ is the spacing between planes in a crystal — so the crystal acts as a diffraction grating for neutrons, exactly as in the Davisson–Germer experiment for electrons.

WHAT: confinement sets a minimum momentum spread $\Delta p\approx \hslash /(2\Delta x)$, using the reduced constant $\hslash =h/2\pi$. WHY $\hslash /(2\Delta x)$: the uncertainty relation $\Delta x\Delta p\ge \hslash /2$ says squeezing position ($\Delta x$ small) forces momentum spread up; taking equality gives the minimum $\Delta p$. $\Delta p=\frac{1.055\times 10^{-34}}{2(1.0\times 10^{-10})}=5.27\times 10^{-25}\text{kg⋅m/s}.$ Taking $p\sim \Delta p$: $\lambda =\frac{h}{p}=\frac{6.626\times 10^{-34}}{5.27\times 10^{-25}}=1.26\times 10^{-9}\text{m}\approx 12.6\text{A˚}.$ Why sensible: the wavelength comes out comparable to (a few times) the box size — a wave just barely fits, which is the physical meaning of being confined. The two frameworks agree in scale.

WHAT & WHY the relativistic form: the electron's rest energy is $m_e{c}^2=8.2\times 10^{-14}\text{J}=511\text{keV}$. Here $K=100\text{keV}$ is a fifth of that — not tiny — so $p=mv$ under-counts the true momentum. The exact relation is $E^2=(pc{)}^2+(m{c}^2{)}^2$ with total energy $E=K+m{c}^2$, giving $pc=\sqrt{(K+m{c}^2{)}^2-(m{c}^2{)}^2}=\sqrt{K^2+2Km{c}^2}.$

(a) Non-relativistic (wrong here, done for contrast): $\lambda =\frac{12.27}{\sqrt{10^5}}=\frac{12.27}{316.2}=0.0388\text{A˚}=3.88\times 10^{-12}\text{m}.$

(b) Relativistic: with $K=1.6\times 10^{-14}\text{J}$, $m{c}^2=8.2\times 10^{-14}\text{J}$: $K^2=2.56\times 10^{-28},2Km{c}^2=2(1.6\times 10^{-14})(8.2\times 10^{-14})=2.62\times 10^{-27}.$ $pc=\sqrt{2.56\times 10^{-28}+2.62\times 10^{-27}}=\sqrt{2.88\times 10^{-27}}=5.37\times 10^{-14}\text{J},$ $p=\frac{5.37\times 10^{-14}}{3\times 10^8}=1.79\times 10^{-22}\text{kg⋅m/s},\lambda =\frac{6.626\times 10^{-34}}{1.79\times 10^{-22}}=3.70\times 10^{-12}\text{m}=0.0370\text{A˚}.$

(c) Compare: non-relativistic gives $0.0388\text{A˚}$; relativistic gives $0.0370\text{A˚}$ — the shortcut is about $5\%$ too long. At higher voltages the error grows fast.

WHAT: equal $\lambda$ ⇒ equal momentum $p=h/\lambda$, but energy depends on how each stores it. WHY split into two cases: a photon's energy is $pc$ (linear in $p$) while a slow electron's is $p^2/2m$ — the same $p$ feeds two different energy formulas. $p=\frac{6.626\times 10^{-34}}{1.0\times 10^{-10}}=6.626\times 10^{-24}\text{kg⋅m/s}.$ Photon: $E_{\gamma }=pc=(6.626\times 10^{-24})(3\times 10^8)=1.99\times 10^{-15}\text{J}=12.4\text{keV}$. Electron (non-relativistic, $K=p^2/2m$): $K_e=\frac{(6.626\times 10^{-24}{)}^2}{2(9.11\times 10^{-31})}=2.41\times 10^{-17}\text{J}=150\text{eV}$. Comment: at the same wavelength the photon carries far more energy ($\sim 12\text{keV}$ vs $\sim 150\text{eV}$), because a photon's energy is $pc$ (linear in $p$) while a slow electron's is $p^2/2m$ (and $c$ is enormous).

WHAT: invert the electron shortcut $\lambda =12.27/\sqrt{V}$ (with $\lambda$ in Å). WHY invert this one: we want the electron voltage that yields a target $\lambda$, exactly what $12.27/\sqrt{V}$ encodes — so solve it for $V$. $\sqrt{V}=\frac{12.27}{{\lambda }_{\text{A˚}}}=\frac{12.27}{0.05}=245.4\Rightarrow V=245.4^2=6.02\times 10^4\text{V}\approx 60\text{kV}.$ Relativity check: $K=eV=60\text{keV}$, again a chunk of the $511\text{keV}$ rest energy, so the true voltage needed is somewhat lower than $60\text{kV}$ (the relativistic momentum is larger for a given $K$). Real electron microscopes indeed run at $\sim 60$–$300\text{kV}$ and use the relativistic formula for precision.

$\lambda =h/p$ is the whole game ::: get $p$ correctly first, then divide $h$ by it. Same speed vs same energy give different ratios ::: $1/m$ (speed) vs $1/\sqrt{m}$ (energy). Bohr orbit ::: whole number of ripples fits, $2\pi r=n\lambda$. What is $\hslash$? ::: the reduced Planck constant $\hslash =h/2\pi =1.055\times 10^{-34}$ J·s. When to go relativistic ::: when $K$ is not $\ll m{c}^2$ ($=511\text{keV}$ for an electron).