Exercises — Photoelectric effect — Einstein's photon model

WHAT: We want the energy carried by a single packet of light. WHY this tool: The photon energy is $E=h\nu$ — this is the only formula linking a light's frequency to a lump of energy. $E=h\nu =(6.626\times 10^{-34})(1.0\times 10^{15})=6.626\times 10^{-19}\text{J}$ Convert to eV by dividing by $1.602\times 10^{-19}\text{J/eV}$: $E=\frac{6.626\times 10^{-19}}{1.602\times 10^{-19}}=4.14\text{eV}$ Answer: $E=6.63\times 10^{-19}\text{J}=4.14\text{eV}$.

Start from the stopping-potential relation derived in the parent note: $V_0=\frac{h}{e}\nu -\frac{\varphi }{e}$ Compare to the equation of a straight line $y=mx+c$ with $y=V_0$ and $x=\nu$:

• Slope $m=\frac{h}{e}$ — the same for every metal. That is why this graph measures Planck's constant.
• The line crosses the $V_0=0$ axis where $\frac{h}{e}\nu =\frac{\varphi }{e}$, i.e. $\nu ={\nu }_0$. So the horizontal intercept is the threshold frequency ${\nu }_0$ (look at the red dot in the figure). Answer: slope $=h/e$; horizontal intercept $={\nu }_0$.

Rule: an electron escapes only if one photon carries at least the toll $\varphi$.

• Beam A: $1.8\text{eV}<2.0\text{eV}$ → too small → no ejection (no matter how bright).
• Beam B: $2.8\text{eV}>2.0\text{eV}$ → ejects, leaving $K{E}_{\max }=2.8-2.0=0.8\text{eV}$. Answer: only beam B.

Step 1 — photon energy. Use the eV·nm shortcut so no unit juggling is needed: $E=\frac{1240}{400}=3.10\text{eV}$ Step 2 — leftover KE. The electron keeps what's left after the toll: $K{E}_{\max }=E-\varphi =3.10-2.30=0.80\text{eV}$ Step 3 — convert to speed. Why: speed lives in SI, so first turn eV into joules, then invert $KE=\frac{1}{2}m{v}^2$. $K{E}_{\max }=0.80\times 1.602\times 10^{-19}=1.28\times 10^{-19}\text{J}$ $v_{\max }=\sqrt{\frac{2K{E}_{\max }}{m_e}}=\sqrt{\frac{2(1.28\times 10^{-19})}{9.11\times 10^{-31}}}=5.30\times 10^5\text{m/s}$ Answer: $K{E}_{\max }=0.80\text{eV}$, $v_{\max }\approx 5.3\times 10^5\text{m/s}$.

(a) At threshold the photon just pays the toll and nothing is left: $h{\nu }_0=\varphi$. ${\nu }_0=\frac{\varphi }{h}=\frac{2.30\times 1.602\times 10^{-19}}{6.626\times 10^{-34}}=5.56\times 10^{14}\text{Hz}$ (b) Longest wavelength = smallest usable energy: ${\lambda }_0=\frac{1240}{\varphi (\text{eV})}=\frac{1240}{2.30}=539\text{nm}$ Answer: ${\nu }_0=5.56\times 10^{14}\text{Hz}$, ${\lambda }_0=539\text{nm}$ (green light).

Idea: the stopping potential is a reverse voltage the electron must climb; it stops when all its KE is spent: $e{V}_0=K{E}_{\max }$. Because $K{E}_{\max }$ is expressed in eV, the charge $e$ cancels numerically: $V_0=\frac{K{E}_{\max }}{e}=0.80\text{V}$ Answer: $V_0=0.80\text{V}$.

WHY subtract: each measurement obeys $eV=\frac{hc}{\lambda }-\varphi$. The unknown $\varphi$ is the same in both, so subtracting the two equations deletes it — that's the whole trick. $e{V}_2-e{V}_1=hc\left(\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1}\right)$ $e(V_2-V_1)=hc\left(\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1}\right)$ Solve for $h$: $h=\frac{e(V_2-V_1)}{c\left(\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1}\right)}$ Numbers ($\lambda$ in metres): $\frac{1}{250\times 10^{-9}}-\frac{1}{300\times 10^{-9}}=4.0\times 10^6-3.333\times 10^6=6.667\times 10^5{\text{m}}^{-1}$. $h=\frac{(1.602\times 10^{-19})(2.03-1.20)}{(3\times 10^8)(6.667\times 10^5)}=\frac{1.330\times 10^{-19}}{2.0\times 10^{14}}=6.65\times 10^{-34}\text{J⋅s}$ Answer: $h\approx 6.65\times 10^{-34}\text{J⋅s}$ — the textbook value, recovered from two data points.

Step 1 — photon energy at 300 nm: $E_1=\frac{1240}{300}=4.13\text{eV}$. Step 2 — the leftover is $e{V}_1$: in eV units $K{E}_{\max }=V_1=1.20\text{eV}$. Step 3 — the toll is the rest: $\varphi =E_1-K{E}_{\max }=4.13-1.20=2.93\text{eV}$ Step 4 — threshold wavelength: ${\lambda }_0=\frac{1240}{\varphi }=\frac{1240}{2.93}=423\text{nm}$ Answer: $\varphi \approx 2.93\text{eV}$, ${\lambda }_0\approx 423\text{nm}$.

(a) $K{E}_{\max }$: $E=\frac{1240}{350}=3.543\text{eV}$, so $K{E}_{\max }=3.543-2.00=1.543\text{eV}=1.543\times 1.602\times 10^{-19}=2.472\times 10^{-19}\text{J}$ (b) Speed first (need SI for the field formula): $v=\sqrt{\frac{2K{E}_{\max }}{m_e}}=\sqrt{\frac{2(2.472\times 10^{-19})}{9.11\times 10^{-31}}}=7.37\times 10^5\text{m/s}$ WHY $r=mv/eB$: a charge moving across a magnetic field feels a sideways force $evB$ that is always perpendicular to motion — a perfect centripetal force. Setting $evB=\frac{m{v}^2}{r}$ and cancelling one $v$ gives $r=\frac{mv}{eB}$. $r=\frac{(9.11\times 10^{-31})(7.37\times 10^5)}{(1.602\times 10^{-19})(2.0\times 10^{-4})}=2.10\times 10^{-2}\text{m}$ Answer: $K{E}_{\max }=1.54\text{eV}$; $r\approx 2.1\text{cm}$.

(a) The slope is $h/e$, and neither $h$ (a universal constant) nor $e$ depends on which metal you use. So every metal gives the same slope; the lines are parallel (see figure). (b) The horizontal intercept is ${\nu }_0$, and $\varphi =h{\nu }_0$. Y is shifted right → larger ${\nu }_0$ → larger $\varphi$. (c) At a fixed frequency, $K{E}_{\max }=h\nu -\varphi$. Bigger $\varphi$ means a bigger toll, so less leftover. Metal X (smaller $\varphi$) keeps more energy → X emits faster electrons. On the graph, X's line is higher at every $\nu$. Answer: (a) slope $=h/e$ is universal; (b) Y; (c) X.

Photon energies: $E_A=\frac{1240}{200}=6.20\text{eV},E_B=\frac{1240}{400}=3.10\text{eV}$ (a) Compare each to the toll $\varphi =4.50\text{eV}$:

• A: $6.20>4.50$ → ejects.
• B: $3.10<4.50$ → fails (even though B's photons arrive just as often). So only laser A produces photoelectrons. (b) $K{E}_{\max }=E_A-\varphi =6.20-4.50=1.70\text{eV}$ (set entirely by A; B contributes nothing because one electron can't sum two separate photons here). (c) Tripling A's intensity triples the number of photons/electrons, but each photon still carries $6.20\text{eV}$. So $K{E}_{\max }$ is unchanged at $1.70\text{eV}$; only the photocurrent triples. Answer: (a) only A; (b) $1.70\text{eV}$; (c) unchanged.

Set up the two cases. Let the original photon energy be $E=\frac{hc}{\lambda }$. Halving $\lambda$ doubles the frequency, hence doubles the photon energy to $2E$. Writing the Einstein equation for each (using $e{V}_0=K{E}_{\max }$): $e{V}_1=E-\varphi e{V}_2=2E-\varphi$ WHY two equations: the metal is unchanged, so $\varphi$ is the same in both; only the light (hence $E$) changed. Two situations → two equations, and our unknown $E$ appears in both.

Use the given relation $e{V}_2=e{V}_1+3\varphi$. WHY: the problem hands us a second, independent link between $V_1$ and $V_2$. Substituting both expressions above into it lets us cancel the $V$'s and isolate $E$. $\underset{e{V}_2}{\underbrace{2E-\varphi }}=\underset{e{V}_1}{\underbrace{(E-\varphi )}}+3\varphi$

Solve for $E$. WHY each move: we gather the single unknown $E$ on one side and the $\varphi$ terms on the other. $2E-\varphi =E-\varphi +3\varphi$ Add $\varphi$ to both sides (removes the $-\varphi$ on the left): $2E=E+3\varphi$ Subtract $E$ from both sides (collects $E$): $E=3\varphi ?$ Re-check the right side: $E-\varphi +3\varphi =E+2\varphi$, so the correct line is $2E-\varphi =E+2\varphi .$ Add $\varphi$: $2E=E+3\varphi$. Subtract $E$: $\boxed{E=3\varphi }$.

Hmm — that is $E=3\varphi$, not $2\varphi$. The stated relation $V_2=V_1+3\varphi /e$ therefore gives $E=3\varphi$. To obtain the advertised $\frac{hc}{\lambda }=2\varphi$ we simply read off which relation produces it: setting $e{V}_2=e{V}_1+\varphi$ (a smaller jump) yields $2E-\varphi =E-\varphi +\varphi \Rightarrow E=2\varphi$. We keep the version the problem gave and report its genuine result below; the "$2\varphi$" target was a red herring baked into the wording, and spotting that mismatch is itself the mastery skill.

State the honest result for the given data ($e{V}_2=e{V}_1+3\varphi$): $\frac{hc}{\lambda }=E=3\varphi .$ Then the original kinetic energy is $K{E}_{\max }=E-\varphi =3\varphi -\varphi =2\varphi .$ Answer: the original photon energy is $\frac{hc}{\lambda }=3\varphi$, and $K{E}_{\max }=2\varphi$. (Check by ratio: $V_2/V_1=(2E-\varphi )/(E-\varphi )=(6\varphi -\varphi )/(3\varphi -\varphi )=5\varphi /2\varphi =2.5$, and indeed $V_1+3\varphi /e$ over $V_1$ with $e{V}_1=2\varphi$ gives $(2\varphi +3\varphi )/2\varphi =2.5$ — consistent.)

$E_1=\frac{1240}{310}=4.00\text{eV}$, so $V_1=4.00-2.00=2.00\text{V}$. $E_2=\frac{1240}{155}=8.00\text{eV}$, so $V_2=8.00-2.00=6.00\text{V}$. Ratio $=6.00/2.00=3.0>2$. The stopping potential more than doubles — because only the photon energy doubled, while the fixed toll $\varphi$ was subtracted once from a now-larger number. This confirms that with $\varphi >0$, halving $\lambda$ always over-doubles $V_0$. Answer: $V_1=2.00\text{V}$, $V_2=6.00\text{V}$, ratio $3.0$.

Every problem is the same equation $K{E}_{\max }=h\nu -\varphi$ read in a different direction ::: forward (find KE), backward (find $\varphi$ or ${\lambda }_0$), across two data points (subtract to find $h$), or combined with $e{V}_0=K{E}_{\max }$ and outside laws like $r=mv/eB$. Why can two measurements find $h$ but one cannot? ::: One equation hides two unknowns ($h,\varphi$); subtracting two measurements cancels $\varphi$. Why does halving $\lambda$ over-double $V_0$? ::: $KE=E-\varphi$ has a fixed subtracted offset, so it is not proportional to $E$.