ARCHETYPE: Photoelectric-Effect Numericals

Hand an examiner one photocell, one metal and one lamp, and they can mint a dozen different questions from it: find the work function from a threshold wavelength, the stopping potential from a colour of light, the maximum electron speed, the metal's identity from two data points, the number of electrons a milliwatt beam kicks out per second. Every one of them is the same physics — Einstein's one-line equation, energy in equals energy to escape plus energy left over — dressed in different exam clothes. Learn the machine once and the whole family becomes fill-in-the-blanks. This is a numerical gym, so the bookkeeping is the article. At its centre sits the equation you met in the concept article: a photon of frequency $f$ carries energy $hf$ ; it spends the work function $\varphi$ prising one electron off the surface, and whatever is left over becomes the electron's maximum kinetic energy, $KE_{\max} = hf - \varphi$ . Stop those fastest electrons with a reverse voltage and that voltage — the stopping potential $V_0$ — reads the leftover energy directly, because $eV_0 = KE_{\max}$ . Three symbols, one straight line, and every photoelectric numerical on CBSE, ICSE, JEE, NEET, AP Physics 2 and A-Level 9702 falls out of it. The trap that costs half the class its marks is a single confusion: brightness versus colour. Intensity sets how many electrons come out (the current); frequency sets how fast they come out (the stopping potential). Keep those two channels separate, carry the photon shortcut $E[\text{eV}] = 1240/\lambda[\text{nm}]$ on the tip of your pen, and the questions turn from feared to routine. This article drills the master recipe and the variants examiners actually pose, with every number computed to the last digit.

Key takeaways

The engine of every photoelectric numerical is Einstein's equation $KE_{\max} = hf - \varphi = h(f - f_0)$ : photon energy in, work function spent, kinetic energy left over.
The stopping potential reads the leftover energy directly: $eV_0 = KE_{\max}$ , so numerically $V_0$ in volts equals $KE_{\max}$ in electronvolts.
Use the photon shortcut $E[\text{eV}] = 1240/\lambda[\text{nm}]$ — a 200 nm photon is $6.20$ eV, a 400 nm photon is $3.10$ eV — and never touch $h$ and $c$ separately unless the question forces SI.
Threshold is where the leftover runs out: $f_0 = \varphi/h$ , $\lambda_0 = 1240/\varphi[\text{eV}]$ nm. Below $f_0$ (above $\lambda_0$ ) nothing comes out, however bright the lamp.
Two colours give two stopping potentials, and the slope of the $V_0$ -versus- $f$ line is $h/e = 4.14\times10^{-15}$ V·s — this is exactly how Millikan measured Planck's constant.
Intensity sets the current, frequency sets the energy — the saturation current counts electrons, the stopping potential clocks their speed.
Key idea: one equation, $eV_0 = hf - \varphi$ , plus the threshold $\lambda_0 = 1240/\varphi$ and the slope $h/e$ , solve the entire photoelectric numerical family — provided you never let a brighter lamp change the electron energy.

The core insight

Every photoelectric numerical reduces to Einstein's energy balance $eV_0 = KE_{\max} = hf - \varphi$ : the photon energy $hf$ (or $1240/\lambda$ in eV) minus the work function $\varphi$ gives the maximum kinetic energy, which the stopping potential measures directly; threshold is where that leftover vanishes ( $\lambda_0 = 1240/\varphi$ ) and the $V_0$ -versus- $f$ slope is the constant $h/e$ .

Where this lives today

The arithmetic you just drilled is the operating manual for a whole industry of light detectors. Every photomultiplier tube — the workhorse of medical PET scanners, particle-physics calorimeters and night-vision optics — begins with a photocathode obeying $KE_{\max} = hf - \varphi$ , and engineers choose its material by exactly the work-function-versus-threshold logic of example 2: a low- $\varphi$ alkali cathode to catch visible light, a wider-gap material to see only the ultraviolet. Solar-cell physicists run the same threshold calculation for the band gap that sets which colours a panel can harvest. Photoelectron spectroscopy — the standard tool for reading the electronic structure of surfaces and catalysts — is quite literally the two-colour method of example 3 turned into a measuring instrument: shine monochromatic light of known energy, measure the electrons' kinetic energy, and read off the binding energies inside the material. The stopping-potential graph Millikan drew in 1916 is now printed on the calibration certificate of instruments in every surface-science lab in the world.

The setup

Bookkeeping first, because in a numerical gym the accounting is the physics. Light of frequency $f$ falls on a clean metal surface whose electrons are held in by a binding energy $\varphi$ — the work function, the minimum energy needed to lift the least-bound electron out through the surface. In the photon picture the light arrives in indivisible lumps of energy $hf$ , and one electron absorbs one whole lump. That electron must first pay $\varphi$ to escape; if the lump is smaller than $\varphi$ nothing happens, and if it is larger, the surplus rides out with the electron as kinetic energy. The most energetic electrons — the ones that were least tightly bound and lose nothing extra on the way out — carry

KE_{\max} = hf - \varphi. \qquad (1)

where: $KE_{\max}$ — the maximum kinetic energy of the ejected photoelectrons; $hf$ — the energy of one photon ( $h = 6.63\times10^{-34}$ J·s is Planck's constant); $\varphi$ — the work function of the metal. Read it as a cash-flow statement: income $hf$ , fixed cost $\varphi$ , profit $KE_{\max}$ . If income falls below the fixed cost the electron never leaves. To measure $KE_{\max}$ you do not chase the electrons with a stopwatch — you stop them with a voltage. Make the collector plate negative so it repels the electrons; wind the voltage up until even the fastest electron is turned back and the current just falls to zero. That critical voltage is the stopping potential $V_0$ , and the work done against it exactly drains the fastest electron's kinetic energy:

eV_0 = KE_{\max} = hf - \varphi. \qquad (2)

where: $V_0$ — the stopping (cut-off) potential; $e = 1.60\times10^{-19}$ C — the electron charge. Because energy in electronvolts is charge-times-volts by definition, equation (2) has a gift built in: $V_0$ measured in volts is numerically equal to $KE_{\max}$ in electronvolts. A leftover of $1.7$ eV means a stopping potential of exactly $1.7$ V. That is why exam problems and this gym live in electronvolts — the arithmetic is almost done before you start.

Figure

The photon energy budget for 200 nm ultraviolet light on tungsten with the 6.2 electronvolt photon split into a 4.5 electronvolt escape cost and a 1.7 electronvolt kinetic surplus read off by the stopping potential

The master relations

Three tools run the entire family. The first is the photon energy in exam-friendly units. Starting from $E = hf = hc/\lambda$ and folding in $h$ , $c$ and $e$ , the product $hc$ comes out as $1240$ eV·nm to three figures, so

E[\text{eV}] = \frac{1240}{\lambda[\text{nm}]}. \qquad (3)

where: $E$ — the photon energy in electronvolts; $\lambda$ — the wavelength in nanometres. This single line replaces three constants: a $400$ nm photon is $1240/400 = 3.10$ eV, a $500$ nm photon is $2.48$ eV, a $656$ nm (red) photon is $1.89$ eV. Commit $1240$ to memory and most board problems need no calculator gymnastics at all. The second tool is the threshold, the boundary where the leftover $KE_{\max}$ falls to zero. Setting $hf = \varphi$ in equation (1) gives the threshold frequency, and translating to wavelength with equation (3):

f_0 = \frac{\varphi}{h}, \qquad \lambda_0 = \frac{1240}{\varphi[\text{eV}]}\ \text{nm}. \qquad (4)

where: $f_0$ — the threshold (cut-off) frequency; $\lambda_0$ — the threshold wavelength. Photons below $f_0$ (equivalently, wavelengths longer than $\lambda_0$ ) cannot free a single electron no matter how intense the beam. For caesium ( $\varphi = 2.14$ eV) the threshold is $\lambda_0 = 1240/2.14 = 579$ nm — yellow light works, red does not; for tungsten ( $\varphi = 4.5$ eV) it is a deep-ultraviolet $276$ nm. Note the inverse: a bigger work function means a shorter threshold wavelength. The third tool is the straight line. Rearranging equation (2) as $V_0 = (h/e)f - \varphi/e$ is the equation of a line in the variables $V_0$ and $f$ : gradient $h/e$ , horizontal intercept at $f_0$ , vertical intercept at $-\varphi/e$ .

V_0 = \frac{h}{e}\,f - \frac{\varphi}{e}, \qquad \text{slope} = \frac{h}{e} = 4.14\times10^{-15}\ \text{V}\,\cdot\,\text{s}. \qquad (5)

where: the slope $h/e$ is a universal constant, the same for every metal — only the intercepts move as the metal changes. This is the deepest single fact in the topic: measure the slope of anyone's $V_0$ -versus- $f$ graph and you have measured Planck's constant. Millikan did exactly this in 1916.

Figure

Straight-line graph of stopping potential against frequency for caesium a rising line of universal slope h over e crossing the frequency axis at the threshold and extrapolating to a negative intercept equal to minus the work function over e

The variants examiners actually pose

Find the work function. Given a threshold wavelength $\lambda_0$ , invert equation (4): $\varphi = 1240/\lambda_0[\text{nm}]$ eV. Given instead a stopping potential at a known wavelength, rearrange equation (2): $\varphi = E - eV_0 = 1240/\lambda - V_0$ (with $V_0$ in volts read as eV). Both are one line of algebra.

Find the stopping potential. The workhorse: $V_0 = E - \varphi = 1240/\lambda - \varphi$ , in volts. If the answer comes out negative or zero, the light is below threshold and no electrons are emitted — a favourite trick.

Find Planck's constant (two-colour method). Shine two different wavelengths, measure two stopping potentials, and take the slope: $h/e = (V_{0,2} - V_{0,1})/(f_2 - f_1)$ . Multiply by $e$ to recover $h$ . The work function drops out of the difference, so you get $h$ without ever knowing $\varphi$ .

Count the electrons (power and efficiency). Intensity questions live in a separate channel. A beam of power $P$ at wavelength $\lambda$ delivers $P/E_{\text{photon}}$ photons per second; a fraction $\eta$ (the quantum efficiency) eject electrons, giving $n = \eta P/E_{\text{photon}}$ photoelectrons per second, and a saturation current $I_{\text{sat}} = ne$ . Notice: this branch uses the photon energy only to count photons — it never touches $\varphi$ or $V_0$ , because how many electrons come out has nothing to do with how fast they go.

Derivation AThe master recipe: stopping potential from wavelength and work function

The goal is to turn the three given quantities of almost every photoelectric problem — a wavelength $\lambda$ , a work function $\varphi$ , and the constants — into the stopping potential $V_0$ , and to see exactly where each input enters so the recipe can be run backwards for any unknown. The strategy is pure energy bookkeeping and algebra: no calculus, no differential equations, just Einstein's one-line balance rearranged. Watch the units travel: electronvolts throughout, so the final volts fall out for free.

Step 1. Convert the light to photon energy. One photon of wavelength $\lambda$ carries $E = hc/\lambda$ , which in exam units is $E = 1240/\lambda[\text{nm}]$ eV by equation (3). For $\lambda = 200$ nm this is $E = 1240/200 = 6.20$ eV.

Step 2. Subtract the escape cost. By equation (1) the fastest electron keeps $KE_{\max} = E - \varphi$ . For tungsten, $\varphi = 4.5$ eV, so $KE_{\max} = 6.20 - 4.5 = 1.70$ eV. If this were negative you would stop here and report "no emission".

Step 3. Read the leftover as a voltage. Equation (2) says $eV_0 = KE_{\max}$ , and because one electronvolt is one electron-charge times one volt, the number of electronvolts of kinetic energy equals the number of volts of stopping potential: $V_0 = 1.70$ V.

Step 4. Run it backwards for any unknown. Given $V_0$ and $\lambda$ , the same chain gives $\varphi = 1240/\lambda - V_0$ . Given $V_0$ and $\varphi$ , it gives $\lambda = 1240/(V_0 + \varphi)$ . The recipe is one equation solved for whichever letter the question hides. Verified: $E = 6.20$ eV, $KE_{\max} = 1.70$ eV and $V_0 = 1.70$ V are recomputed independently in verify_hs2-photoelectric-numericals.py.

Go deeper · Derivation B · 1 minThe two-colour slope method: measuring Planck's constant without knowing the work function

Derivation BThe two-colour slope method: measuring Planck's constant without knowing the work function

The goal is Millikan's trick — extract $h$ (and, as a bonus, $\varphi$ ) from two stopping-potential measurements at two wavelengths, without ever being told the metal's work function. The strategy is to write Einstein's equation twice and subtract: the unknown $\varphi$ is the same in both, so it cancels, leaving a clean ratio. This is algebra of simultaneous equations, nothing more.

Step 1. Write equation (2) at each frequency: $eV_{0,1} = hf_1 - \varphi$ and $eV_{0,2} = hf_2 - \varphi$ , with $f = c/\lambda$ . Take caesium lit first by $\lambda_1 = 400$ nm ( $f_1 = 7.5\times10^{14}$ Hz) then by $\lambda_2 = 300$ nm ( $f_2 = 1.0\times10^{15}$ Hz).

Step 2. Subtract to kill $\varphi$ : $e(V_{0,2} - V_{0,1}) = h(f_2 - f_1)$ . The work function has vanished — the difference of two stopping potentials depends only on the difference of frequencies and on $h$ .

Step 3. Solve for the slope. $\dfrac{h}{e} = \dfrac{V_{0,2} - V_{0,1}}{f_2 - f_1}$ . With the measured $V_{0,1} = 0.96$ V and $V_{0,2} = 1.99$ V, the slope is $(1.99 - 0.96)/(1.0\times10^{15} - 7.5\times10^{14}) = 1.03/2.5\times10^{14} = 4.13\times10^{-15}$ V·s.

Step 4. Recover the constants. Multiply by $e$ : $h = e\times4.13\times10^{-15} = 6.61\times10^{-34}$ J·s — Planck's constant, to within a per cent of the accepted $6.63\times10^{-34}$ , from nothing but two voltmeter readings. Back-substitute one point to get $\varphi = hf_1 - eV_{0,1} = 3.10 - 0.96 = 2.14$ eV, identifying the metal as caesium. Verified: the slope $4.13\times10^{-15}$ V·s, the recovered $h = 6.61\times10^{-34}$ J·s and $\varphi = 2.14$ eV are recomputed in verify_hs2-photoelectric-numericals.py.

Trap table and limiting cases

pitfall or limiting case	what actually happens	the fix
brighter light, faster electrons?	no — intensity raises the current, not $KE_{\max}$	frequency sets energy, intensity sets number
$\lambda > \lambda_0$ (below threshold)	zero emission, however intense	check $E > \varphi$ before computing $V_0$
$\lambda = \lambda_0$ exactly	$KE_{\max} = 0$ , $V_0 = 0$	electrons just barely escape with no speed
$V_0$ depends on the metal?	the value does; the slope $h/e$ does not	only intercepts move between metals
using $h$ and $c$ separately	slow and error-prone	use $E[\text{eV}] = 1240/\lambda[\text{nm}]$
eV versus joule	$V_0$ in volts $=$ $KE_{\max}$ in eV numerically	multiply by $e$ only when SI joules are wanted
$\lambda \to$ very small (hard UV/X-ray)	$KE_{\max} \to hf$ , work function negligible	at high energy $\varphi$ is a tiny correction

Common trapA brighter lamp sends more photons, not bigger ones

The single costliest misconception in the whole topic is believing that turning the lamp up makes the electrons come out faster. It does not. Doubling the intensity at a fixed colour doubles the number of photons per second, so twice as many electrons are ejected — the saturation current doubles — but each photon still carries the same energy $hf$ , so each electron still leaves with the same maximum kinetic energy and the stopping potential $V_0$ is unchanged. To speed the electrons up you must change the colour: raise the frequency (shorten the wavelength), which raises $hf$ and hence $KE_{\max}$ . Two separate channels: intensity feeds the current, frequency feeds the energy. Any exam graph that shows a higher stopping potential is telling you the frequency changed, never the brightness — and a taller saturation plateau at the same $V_0$ is telling you the intensity changed, never the colour.

Worked examples

1 · Board level — the master template (CBSE and NEET pattern)

Ultraviolet light of wavelength $200$ nm falls on a clean tungsten surface of work function $4.5$ eV. Find the energy of each photon, the maximum kinetic energy of the emitted electrons, and the stopping potential.

Step 1. Photon energy by the shortcut, equation (3): $E = 1240/\lambda = 1240/200 = \boxed{6.20\ \text{eV}}$ .

Step 2. Subtract the work function, equation (1): $KE_{\max} = E - \varphi = 6.20 - 4.5 = \boxed{1.70\ \text{eV}}$ . It is positive, so electrons are emitted.

Step 3. The stopping potential reads the leftover directly, equation (2): $eV_0 = KE_{\max}$ , so $V_0 = \boxed{1.70\ \text{V}}$ .

Step 4. Sense-check: the photon (6.2 eV) comfortably clears the 4.5 eV barrier, and $200$ nm is well below tungsten's threshold $\lambda_0 = 1240/4.5 = 276$ nm, so emission is expected. Everything consistent.

2 · Board level — threshold, and does the light even work (ICSE and CBSE pattern)

Caesium has a work function of $2.14$ eV. Find its threshold wavelength and threshold frequency. Then decide whether $400$ nm violet light ejects electrons, and if so give the stopping potential.

Step 1. Threshold wavelength, equation (4): $\lambda_0 = 1240/\varphi = 1240/2.14 = \boxed{579\ \text{nm}}$ (yellow-green).

Step 2. Threshold frequency, equation (4): $f_0 = \varphi/h = (2.14\times1.60\times10^{-19})/(6.63\times10^{-34}) = \boxed{5.16\times10^{14}\ \text{Hz}}$ .

Step 3. Test the $400$ nm light: its photon energy is $E = 1240/400 = 3.10$ eV, which exceeds $\varphi = 2.14$ eV — and $400$ nm is shorter than $\lambda_0 = 579$ nm — so electrons are emitted.

Step 4. Stopping potential: $KE_{\max} = 3.10 - 2.14 = 0.96$ eV, so $V_0 = \boxed{0.96\ \text{V}}$ . Note that a red $656$ nm photon ( $1.89$ eV) would be below threshold and eject nothing.

3 · Competitive level — measure Planck's constant from two colours (JEE-Advanced and A-Level pattern)

A photocell records a stopping potential of $0.96$ V under $400$ nm light and $1.99$ V under $300$ nm light. From these two data points alone, find Planck's constant and the work function of the cathode.

Step 1. Convert to frequencies: $f_1 = c/\lambda_1 = (3.0\times10^{8})/(400\times10^{-9}) = 7.5\times10^{14}$ Hz; $f_2 = (3.0\times10^{8})/(300\times10^{-9}) = 1.0\times10^{15}$ Hz.

Step 2. The work function cancels in the difference of two Einstein equations, leaving the slope: $h/e = (V_{0,2} - V_{0,1})/(f_2 - f_1) = (1.99 - 0.96)/(1.0\times10^{15} - 7.5\times10^{14}) = 1.03/(2.5\times10^{14}) = 4.13\times10^{-15}$ V·s.

Step 3. Recover Planck's constant: $h = e\times(h/e) = (1.60\times10^{-19})(4.13\times10^{-15}) = \boxed{6.61\times10^{-34}\ \text{J}\,\cdot\,\text{s}}$ , within $1\%$ of the accepted $6.63\times10^{-34}$ .

Step 4. Recover the work function from either point: $\varphi = hf_1 - eV_{0,1} = 3.10 - 0.96 = \boxed{2.14\ \text{eV}}$ , so the cathode is caesium. Timing note: about 90 seconds. The whole method is "write it twice and subtract" — the moment you see that $\varphi$ cancels, only the slope remains, and there is no need to solve for the work function first.

4 · Competitive level — count the electrons and the saturation current (JEE-Main and A-Level pattern)

A $1.0$ mW beam of $400$ nm light strikes a photocathode of quantum efficiency $5\%$ (one electron per twenty incident photons). How many photoelectrons are emitted per second, and what saturation current do they carry?

Step 1. Energy of one photon in joules: $E = (1240/400)$ eV $= 3.10$ eV $= 3.10\times1.60\times10^{-19} = 4.96\times10^{-19}$ J.

Step 2. Photons arriving per second: $P/E = (1.0\times10^{-3})/(4.96\times10^{-19}) = 2.02\times10^{15}$ photons per second.

Step 3. Photoelectrons per second at $5\%$ efficiency: $n = 0.05\times2.02\times10^{15} = \boxed{1.01\times10^{14}\ \text{s}^{-1}}$ .

Step 4. Saturation current, $I_{\text{sat}} = ne = (1.01\times10^{14})(1.60\times10^{-19}) = 1.61\times10^{-5}$ A $= \boxed{16.1\ \mu\text{A}}$ . Timing note: about 60 seconds. This branch never touches $\varphi$ or $V_0$ — the photon energy is used only to count photons. Doubling the beam power would double the current and leave the stopping potential untouched, the whole point of the intensity channel.

Exam traps

Confusing intensity with frequency. Intensity sets how many electrons (the saturation current); frequency sets how fast (the stopping potential). Turning up the lamp never raises $V_0$ . This is the single biggest mark-loser in the topic; state which channel each number lives in before you compute.

Forgetting to check threshold. Always confirm $E > \varphi$ (equivalently $\lambda < \lambda_0$ ) before writing a stopping potential. If the photon energy is below the work function, the answer is "no emission", not a negative $V_0$ . Examiners plant sub-threshold light precisely to catch the automatic subtracter.

Splitting $h$ and $c$ when you needn't. For energies, $E[\text{eV}] = 1240/\lambda[\text{nm}]$ is faster and less error-prone than juggling $6.63\times10^{-34}$ , $3\times10^{8}$ and $1.6\times10^{-19}$ . Reserve the SI constants for when the question explicitly wants joules or frequency.

Mixing electronvolts and joules. $V_0$ in volts equals $KE_{\max}$ in electronvolts — do not multiply by $e$ a second time. Convert to joules only when the current or an SI energy is asked for. A stray factor of $e$ (or $10^{19}$ ) is an instant order-of-magnitude blunder.

Unit slips on wavelength and power. Keep nanometres in the $1240$ shortcut, but convert to metres for $f = c/\lambda$ , and milliwatts to watts for photon counting. Carry four significant figures through the working and present two or three.

Board scope. Photoelectric numericals are firmly examined on CBSE, ICSE, JEE-Main, JEE-Advanced and NEET, and also on AP Physics 2 (the Modern Physics unit) and A-Level 9702, which both do stopping-potential and threshold problems and the $h/e$ slope. IB Physics carries the photoelectric equation at Higher Level only and does not set it as a numerical-drill objective, and IGCSE 0625 has no photoelectric content at all — do not revise this lane for an IGCSE paper.

Origins · 1 minRobert Millikan and how this was discovered

Robert Millikan

1868–1953 · Illinois and California, USA

Robert Andrews Millikan was the American experimentalist who nailed down two of the numbers this whole gym runs on. His 1909–1913 oil-drop experiment measured the electron's charge $e$ — the very constant that turns a stopping potential into an energy — to a then-unmatched precision. He then turned, sceptically, to Einstein's 1905 photoelectric equation, which he frankly disbelieved; he spent a decade building meticulous vacuum photocells to disprove it. Instead his stopping-potential-versus-frequency graphs came out dead straight, and in 1916 he read Planck's constant off their slope, $h/e$ , confirming Einstein's quantum picture to about half a per cent. It is a model of science overruling a scientist's prejudice by measurement. Millikan received the Nobel Prize in Physics in 1923 for the elementary charge and the photoelectric effect, and later ran the California Institute of Technology into a research powerhouse. Every slope method in this article is his.

History

1887

— Heinrich Hertz notices that ultraviolet light makes a spark jump more easily across a gap, the first sighting of the photoelectric effect.

1902

— Philipp Lenard shows the ejected electrons' maximum energy is independent of the light's intensity (and rises with its frequency), a result classical wave theory could not explain.

1905

— Albert Einstein proposes light quanta and the equation KE max = hf minus phi, for which he later wins the 1921 Nobel Prize.

1916

— Robert Millikan's precise stopping-potential measurements confirm Einstein's equation and read Planck's constant off the slope h over e.

1923

— Millikan is awarded the Nobel Prize in Physics for the elementary charge and his work on the photoelectric effect.

Going further

Texts. NCERT Physics Class 12, Ch. 11 ("Dual Nature of Radiation and Matter"), the photoelectric sections and exercises · H. C. Verma, Concepts of Physics Vol. 2, "Photoelectric Effect and Wave–Particle Duality," worked problems · D. C. Pandey / Irodov photoelectric problem sets (for the two-colour slope and photon-count variants) · Cambridge A-Level 9702, quantum-physics stopping-potential problems.

Provenance. Standard exam-archive problems. Recompute: master template $\lambda = 200$ nm on W ( $\varphi = 4.5$ ) $\Rightarrow E = 6.20$ eV, $KE_{\max} = 1.70$ eV, $V_0 = 1.70$ V; caesium ( $\varphi = 2.14$ ) threshold $\lambda_0 = 579$ nm, $f_0 = 5.16\times10^{14}$ Hz; $400$ nm on Cs $\Rightarrow E = 3.10$ eV, $V_0 = 0.96$ V; two-colour ( $400$ and $300$ nm, $V_0 = 0.96$ and $1.99$ V) $\Rightarrow$ slope $4.13\times10^{-15}$ V·s, $h = 6.61\times10^{-34}$ J·s, $\varphi = 2.14$ eV; photon count ( $1$ mW, $400$ nm, $5\%$ ) $\Rightarrow 2.02\times10^{15}$ photons/s, $1.01\times10^{14}$ electrons/s, $I_{\text{sat}} = 16.1\ \mu$ A; photon references $E(500\,\text{nm}) = 2.48$ eV, $E(656\,\text{nm}) = 1.89$ eV; slope constant $h/e = 4.14\times10^{-15}$ V·s.

Prerequisites. Einstein's Photoelectric Equation — the concept article that derives $KE_{\max} = hf - \varphi$ and the $V_0$ – $f$ line this gym uses. Leads to. the university seam The Photoelectric Effect and the Photon (`aq-qm-photoelectric-effect-photon`), which develops the quantum picture and the photon momentum in full.

Go deeper (UG). The rigorous quantum-mechanical account — the photon as a quantum of the electromagnetic field, the exact energy distribution of photoelectrons, and the tunnelling and band-structure corrections to the simple work function — is developed in The Photoelectric Effect and the Photon. Enrichment only; nothing here depends on it.

Why is the stopping potential in volts numerically equal to the kinetic energy in electronvolts?

Because of how the electronvolt is defined. One electronvolt is the energy an electron gains crossing a potential difference of one volt: $1\ \text{eV} = e\times1\ \text{V}$ . The stopping condition is $eV_0 = KE_{\max}$ , so if you measure $KE_{\max}$ in electronvolts, the $e$ on both sides cancels and the number of electronvolts equals the number of volts. A leftover kinetic energy of $1.70$ eV is stopped by exactly $1.70$ V. This is not a coincidence or an approximation — it is built into the units, and it is why every photoelectric problem is easiest in electronvolts. Convert to joules only if a question asks for an SI energy or a current, where $KE_{\max}$ in joules is $eV_0$ .

How do I know whether any electrons are emitted at all?

Check the photon energy against the work function before you do anything else. Compute $E = 1240/\lambda[\text{nm}]$ eV and compare with $\varphi$ . If $E > \varphi$ (equivalently, if the wavelength is shorter than the threshold $\lambda_0 = 1240/\varphi$ ), electrons come out and $KE_{\max} = E - \varphi$ . If $E \le \varphi$ , no electrons are emitted, however intense the light, and the correct answer is "zero photocurrent", not a negative stopping potential. Examiners deliberately set light just below threshold to catch students who subtract on autopilot and report an impossible negative $V_0$ . Making this one-line check your first step turns those trick questions into free marks.

The two-colour method gives h without the work function — how can that be?

Because subtraction erases the work function. Write Einstein's equation at each frequency: $eV_{0,1} = hf_1 - \varphi$ and $eV_{0,2} = hf_2 - \varphi$ . The same $\varphi$ sits in both — it is a property of the metal, not of the light — so when you subtract one equation from the other, the $\varphi$ terms cancel and you are left with $e(V_{0,2} - V_{0,1}) = h(f_2 - f_1)$ . The slope of the line therefore carries $h/e$ and nothing about the metal. This is exactly why the $V_0$ -versus- $f$ graph has the same gradient for caesium, sodium and tungsten: only the intercepts, which do contain $\varphi$ , differ. If you later want $\varphi$ , substitute either data point back into a single Einstein equation.

What is the difference between saturation current and stopping potential in a numerical?

They live in two independent channels, and mixing them is the classic error. The saturation current is the maximum photocurrent, reached when the collector is positive enough to gather every ejected electron; its size is set by how many electrons are emitted per second, which is set by the light's intensity (through the photon-counting formula $n = \eta P/E_{\text{photon}}$ ). The stopping potential is the reverse voltage that just halts the fastest electron; its size is set by how fast the electrons are, which is set by the light's frequency (through $eV_0 = hf - \varphi$ ). Intensity moves the saturation plateau up and down; frequency slides the stopping potential left and right. A single question often gives you data for both — keep them in separate columns.

Which value of Planck's constant and which shortcut should I use in exams?

Use $h = 6.63\times10^{-34}$ J·s $= 4.14\times10^{-15}$ eV·s, $c = 3.0\times10^{8}$ m/s and $e = 1.6\times10^{-19}$ C — the standard board values. For photon energies the shortcut $E[\text{eV}] = 1240/\lambda[\text{nm}]$ saves time and error; it is just $hc/e$ rounded to $1240$ eV·nm. When you need a frequency, use $f = c/\lambda$ in SI (metres). The one place the two shortcuts can disagree at the third digit is a two-colour slope, where the energy shortcut and the frequency route are rounded differently — that is why example 3's slope comes out $4.13\times10^{-15}$ rather than exactly $4.14\times10^{-15}$ ; both are "Planck's constant to two figures". Quote the value your board's data sheet prints, and keep four figures in the working.

Is this gym the same as the concept article on Einstein's photoelectric equation?

It shares the physics but not the purpose. The concept article derives the equation $KE_{\max} = hf - \varphi$ — why light must be quantised, why there is a threshold, why classical waves fail — and explains the meaning of every term. This article assumes all that and drills using it fast and correctly under exam time: the master solve, the threshold check, the two-colour slope, the photon count. Think of the concept article as the theory lecture and this as the problem class. If a formula here feels unmotivated, revisit the concept article; if you understand the physics but keep losing marks to intensity-versus-frequency slips or forgotten threshold checks, this gym is where you fix them.