Exercises — Dual nature of matter — de Broglie λ = h - p

WHAT form? We are given mass and speed, so the simplest form is $\lambda =h/(mv)$ — no need for energy versions. $\lambda =\frac{6.626\times 10^{-34}}{(9.11\times 10^{-31})(2.0\times 10^6)}$ Multiply the denominator: $(9.11\times 10^{-31})(2.0\times 10^6)=1.822\times 10^{-24}\text{kg m/s}$ (this is the momentum $p$). $\lambda =\frac{6.626\times 10^{-34}}{1.822\times 10^{-24}}=3.64\times 10^{-10}\text{m}=0.364\text{nm}$ Sanity check: ~Å scale, exactly the size where electrons diffract off crystals.

WHY rearrange? We want $p$, so invert $\lambda =h/p\Rightarrow p=h/\lambda$. Nothing about mass is needed — momentum is what the wavelength directly encodes. $p=\frac{h}{\lambda }=\frac{6.626\times 10^{-34}}{5.0\times 10^{-11}}=1.325\times 10^{-23}\text{kg m/s}$

WHY the shortcut? For an electron the messy constants collapse to $\lambda (\text{nm})\approx 1.226/\sqrt{V}$. It already contains $h$, $m_e$ and $q$. $\lambda =\frac{1.226}{\sqrt{400}}=\frac{1.226}{20}=0.0613\text{nm}=6.13\times 10^{-11}\text{m}$

Step 1 — convert energy to joules. The formula needs SI: $KE=1000\text{eV}\times 1.6\times 10^{-19}=1.6\times 10^{-16}\text{J}$ Step 2 — WHY the $\sqrt{2mKE}$ form? We know energy, not speed, so use $\lambda =h/\sqrt{2mKE}$. $2mKE=2(1.67\times 10^{-27})(1.6\times 10^{-16})=5.344\times 10^{-43}$ $\sqrt{5.344\times 10^{-43}}=7.31\times 10^{-22}\text{kg m/s}$ $\lambda =\frac{6.626\times 10^{-34}}{7.31\times 10^{-22}}=9.06\times 10^{-13}\text{m}$

WHY reason with ratios? We don't need numbers — $\lambda =h/(mv)$, and $h,m$ are fixed, so $\lambda \propto 1/v$. $\frac{{\lambda }_2}{{\lambda }_1}=\frac{v_1}{v_2}=\frac{v_1}{2v_1}=\frac{1}{2}$ The wavelength halves. Faster ⇒ more momentum ⇒ shorter wave.

WHY the KE form? Equal $KE$ is the shared quantity, so use $\lambda =h/\sqrt{2mKE}$. Here $h$ and $KE$ are identical for both, so $\lambda \propto 1/\sqrt{m}$. $\frac{{\lambda }_e}{{\lambda }_p}=\sqrt{\frac{m_p}{m_e}}=\sqrt{\frac{1.67\times 10^{-27}}{9.11\times 10^{-31}}}=\sqrt{1833}\approx 42.8$ The electron's wavelength is ~43× longer. Lighter particle at equal energy ⇒ smaller momentum ⇒ bigger wave.

WHY: $\lambda =1.226/\sqrt{V}$, so $\lambda \propto 1/\sqrt{V}$. To halve $\lambda$ we need $\sqrt{V}$ to double, i.e. $V$ to quadruple ($\times 4$). Check: ${\lambda }_{\text{new}}/{\lambda }_{\text{old}}=\sqrt{V_{\text{old}}/V_{\text{new}}}=\sqrt{1/4}=1/2$. ✓

The idea: a stable orbit must hold a standing wave — the circumference equals a whole number of wavelengths, $2\pi r=n\lambda$. For $n=1$ we expect $2\pi r_1=1\cdot \lambda$. Step 1 — wavelength of the orbiting electron: $\lambda =\frac{h}{m_e{v}_1}=\frac{6.626\times 10^{-34}}{(9.11\times 10^{-31})(2.19\times 10^6)}=3.322\times 10^{-10}\text{m}$ Step 2 — circumference: $2\pi r_1=2\pi (5.29\times 10^{-11})=3.324\times 10^{-10}\text{m}$ They match (to rounding): $2\pi r_1\approx \lambda$. This is why orbits are quantised — only whole-number wave-fits survive; anything else destroys itself by interference.

Photon (uses Planck's Quantum Theory, $E=hc/\lambda$): $E_{\gamma }=\frac{hc}{\lambda }=\frac{(6.626\times 10^{-34})(3.0\times 10^8)}{1.0\times 10^{-10}}=1.988\times 10^{-15}\text{J}\approx 12.4\text{keV}$ Electron (non-relativistic, $KE=p^2/2m$ with $p=h/\lambda$): $p=\frac{h}{\lambda }=\frac{6.626\times 10^{-34}}{1.0\times 10^{-10}}=6.626\times 10^{-24}\text{kg m/s}$ $K{E}_e=\frac{p^2}{2m_e}=\frac{(6.626\times 10^{-24}{)}^2}{2(9.11\times 10^{-31})}=2.41\times 10^{-17}\text{J}\approx 150\text{eV}$ The photon carries far more energy (~12,400 eV vs ~150 eV) even though both share the same wavelength. Reason: for a photon $E=pc$ (speed of light), but for a slow electron $E=p^2/2m$ — a much smaller energy for the same $p$.

Naive shortcut: $\lambda \approx 1.226/\sqrt{V}=1.226/\sqrt{10^6}=1.226\times 10^{-3}\text{nm}=1.226\times 10^{-12}\text{m}$. WHY it's wrong here: at $1\text{MeV}$ the electron's kinetic energy ($10^6$ eV) is larger than its rest energy ($m_e{c}^2=5.11\times 10^5$ eV), so it moves near $c$. The shortcut assumed $KE=\frac{1}{2}m{v}^2$ (non-relativistic), which underestimates momentum at high speed. Real momentum is larger ⇒ real $\lambda$ is smaller than the naive $1.23\times 10^{-12}$ m. (Full relativistic value ≈ $0.87\times 10^{-12}$ m.) Lesson: the shortcut is only trustworthy while $qV\ll m_e{c}^2\approx 511\text{keV}$.

$\lambda =\frac{h}{mv}=\frac{6.626\times 10^{-34}}{(1.0\times 10^{-9})(1.0\times 10^{-3})}=6.626\times 10^{-22}\text{m}$ Even this near-invisible grain has $\lambda \sim 10^{-22}$ m — about $10^{12}$ times smaller than an atomic nucleus. No slit or crystal can be that fine, so its wave nature is permanently hidden. This is the boundary from the Heisenberg Uncertainty Principle: waviness only matters when $\lambda$ is comparable to system size.

Step 1 — kinetic energy: $KE=\frac{3}{2}(1.38\times 10^{-23})(300)=6.21\times 10^{-21}\text{J}$ Step 2 — WHY the $\sqrt{2mKE}$ form: we have energy, not speed. $2mKE=2(1.675\times 10^{-27})(6.21\times 10^{-21})=2.080\times 10^{-47}$ $\sqrt{2mKE}=4.561\times 10^{-24}\text{kg m/s}$ $\lambda =\frac{6.626\times 10^{-34}}{4.561\times 10^{-24}}=1.45\times 10^{-10}\text{m}=0.145\text{nm}$ Why it matters: this ~1.5 Å wavelength is atom-spacing sized, which is exactly why thermal neutrons are used to diffract off crystals — the neutron cousin of the Davisson–Germer Experiment.

Given speed $v$ ::: $\lambda =h/mv$ Given kinetic energy $KE$ ::: $\lambda =h/\sqrt{2mKE}$ Given accelerating voltage $V$ (electron) ::: $\lambda \approx 1.226/\sqrt{V}$ nm Given wavelength, want momentum ::: $p=h/\lambda$ Scaling: halve $\lambda$ by changing voltage ::: raise $V$ by factor 4