Exercises — Photon properties — E = hf, p = h - λ

WHAT we need: energy from frequency → use $E=hf$ (the version that takes $f$ directly). $E=hf=(6.626\times 10^{-34})(6.0\times 10^{14})=3.98\times 10^{-19}\text{J}.$ WHY this form: we were given $f$, so no need to detour through $\lambda$.

WHAT we need: momentum from wavelength → the shortest route is $p=h/\lambda$. Convert first: $400\text{nm}=400\times 10^{-9}\text{m}=4.0\times 10^{-7}\text{m}$. $p=\frac{h}{\lambda }=\frac{6.626\times 10^{-34}}{4.0\times 10^{-7}}=1.66\times 10^{-27}\text{kg⋅m/s}.$

Step 1 — WHAT: given $\lambda$, want energy → use $E=hc/\lambda$. $E=\frac{hc}{\lambda }=\frac{(6.626\times 10^{-34})(3\times 10^8)}{0.10\times 10^{-9}}=\frac{1.988\times 10^{-25}}{1.0\times 10^{-10}}=1.988\times 10^{-15}\text{J}.$ Step 2 — WHY convert: joules this small are unreadable; divide by $1.6\times 10^{-19}$ to get eV. $E=\frac{1.988\times 10^{-15}}{1.6\times 10^{-19}}\approx 1.24\times 10^4\text{eV}=12.4\text{keV}.$ A short wavelength ⇒ high energy — exactly why X-rays penetrate flesh.

WHAT: we are handed $p$ and want $\lambda$ → invert $p=h/\lambda$ into $\lambda =h/p$. $\lambda =\frac{h}{p}=\frac{6.626\times 10^{-34}}{2.2\times 10^{-27}}=3.01\times 10^{-7}\text{m}=301\text{nm}.$ Is it visible? The visible band is roughly $400$–$700\text{nm}$. $301\text{nm}$ is shorter, so it is ultraviolet, just outside what our eyes detect.

Step 1 — energy per photon: $E=hc/\lambda =\frac{1.988\times 10^{-25}}{532\times 10^{-9}}=3.74\times 10^{-19}\text{J}.$ Step 2 — WHY divide: power = (energy per photon)$\times$(photons per second), so $N=P/E$. $N=\frac{P}{E}=\frac{2.0\times 10^{-3}}{3.74\times 10^{-19}}\approx 5.35\times 10^{15}\text{photons/s}.$

Reasoning: both $E$ and $p$ go as $1/\lambda$. So the ratio is just the inverse ratio of wavelengths. $\frac{E_B}{E_A}=\frac{{\lambda }_A}{{\lambda }_B}=\frac{600}{200}=3,\frac{p_B}{p_A}=\frac{600}{200}=3.$ Photon B (shorter $\lambda$) has 3× the energy and 3× the momentum. Scaling both the same way is the fingerprint of $E=pc$.

WHAT: ratio of energies of two photons → $E\propto 1/\lambda$, so factor $={\lambda }_{\text{radio}}/{\lambda }_{\text{gamma}}$. $\frac{E_{\gamma }}{E_{\text{radio}}}=\frac{3.0}{1.0\times 10^{-12}}=3.0\times 10^{12}.$ The gamma photon is three trillion times more energetic — same equation, opposite ends of the spectrum.

WHY this tool: the photon delivers all its energy $E$ in one lump; the electron spends $\varphi$ escaping and keeps the rest as kinetic energy. This is Einstein's photoelectric equation $K_{\max }=E-\varphi$ — see Photoelectric Effect. $K_{\max }=E-\varphi =2.48-2.0=0.48\text{eV}.$ If $E<\varphi$, no electron escapes at all — the lump simply isn't big enough, no matter how many lumps arrive.

Algebra (WHY): start from the parent's two definitions and eliminate $\lambda$. $E=\frac{hc}{\lambda },p=\frac{h}{\lambda }⟹\frac{E}{p}=\frac{hc/\lambda }{h/\lambda }=c⟹E=pc.$ This is the same result the Relativistic Energy-Momentum Relation gives when you set rest mass $m_0=0$. Numeric check ($\lambda =500\text{nm}$): $E=\frac{1.988\times 10^{-25}}{5.0\times 10^{-7}}=3.98\times 10^{-19}\text{J}$, and $p=\frac{6.626\times 10^{-34}}{5.0\times 10^{-7}}=1.33\times 10^{-27}\text{kg⋅m/s}$. Then $pc=(1.33\times 10^{-27})(3\times 10^8)=3.98\times 10^{-19}\text{J}=E.$ ✓

WHY momentum enters: force = rate of momentum delivered. Each photon carries $p=E/c$. In one second the beam delivers energy $P\times 1\text{s}$, hence momentum $P/c$ per second. $F=\frac{\Delta p}{\Delta t}=\frac{P}{c}=\frac{1.0}{3\times 10^8}=3.33\times 10^{-9}\text{N}.$ Tiny, but real — this is Radiation Pressure, the force that pushes solar sails. (A reflecting panel would double this to $2P/c$ because momentum reverses.)

Key insight: the de Broglie Wavelength relation $p=h/\lambda$ is the same relation for the electron as for the photon — de Broglie ran the photon rule backwards for matter. $p_{\text{photon}}=\frac{h}{\lambda }=\frac{6.626\times 10^{-34}}{1.0\times 10^{-10}}=6.63\times 10^{-24}\text{kg⋅m/s}.$ Since both share the same $\lambda$, they share the same momentum: $p_{\text{electron}}=6.63\times 10^{-24}\text{kg⋅m/s}$ too. Their energies differ wildly (electron has rest mass, photon does not), but $p$ is fixed purely by $\lambda$.

(a) Photons per second. Energy per photon: $E=hc/\lambda =\frac{1.988\times 10^{-25}}{550\times 10^{-9}}=3.61\times 10^{-19}\text{J}.$ $N=\frac{P}{E}=\frac{100}{3.61\times 10^{-19}}\approx 2.77\times 10^{20}\text{photons/s}.$ (b) Force on the mirror. Power reaching mirror: $P^{\prime }=0.010\times 100=1.0\text{W}$. Reflection ⇒ momentum reverses ⇒ force $=2P^{\prime }/c$. $F=\frac{2P^{\prime }}{c}=\frac{2(1.0)}{3\times 10^8}=6.67\times 10^{-9}\text{N}.$ The $2.0\text{m}$ distance is a decoy — it does not enter, because we were told the fraction of light caught directly.

This is the Compton Effect: a photon behaving like a solid ball that loses energy to a recoiling electron. (a) Energy lost by photon = energy gained by electron. Compute both photon energies via $E=hc/\lambda$. $E=\frac{1.988\times 10^{-25}}{0.0700\times 10^{-9}}=2.840\times 10^{-15}\text{J},$ $E^{\prime }=\frac{1.988\times 10^{-25}}{0.0724\times 10^{-9}}=2.746\times 10^{-15}\text{J}.$ Electron's gain $=E-E^{\prime }=2.840\times 10^{-15}-2.746\times 10^{-15}=9.42\times 10^{-17}\text{J}\approx 589\text{eV}.$ (b) The proof: the photon changed wavelength (hence momentum $p=h/\lambda$). A pure wave bouncing off would keep its wavelength. A wavelength shift means the photon carried and traded momentum like a particle — momentum is the smoking gun.

Give me $\lambda$, I get $E$ how? ::: $E=hc/\lambda$. Give me $\lambda$, I get $p$ how? ::: $p=h/\lambda$. Force of an absorbed beam of power $P$? ::: $F=P/c$. Force of a reflected beam of power $P$? ::: $F=2P/c$. Photons per second from power $P$, wavelength $\lambda$? ::: $N=P/(hc/\lambda )$. Shorter wavelength means ___ energy and ___ momentum. ::: more; more (both $\propto 1/\lambda$).