Dual Nature: theory | Physics Circle

CONCEPT

Discharge, Cathode Rays & Thomson

In one line
At low pressure and high voltage a gas conducts, and the cathode gives out fast electrons called cathode rays.

DISCHARGE

25.1.1 Discharge through gases

At normal pressure a gas is a poor conductor; about 30 kV is needed for a spark across a few cm of air.
The tube is about 30 cm to 40 cm long and 4 cm in diameter.
At low pressure the mean free path is larger, so electrons gain more KE between collisions.
Fast electrons hit neutral atoms and ionise them.
Coloured glow: charged particles from the cathode collide with gas atoms, and the excited atoms give out light.

Discharge tube with a vacuum pump

Pressure (mm of Hg)	What is seen
10	zig-zag spark, cracking sound
4	glow at the electrodes only (dark discharge)
below 4	positive column fills the tube
1.65	negative glow; Faraday’s dark space
0.8	Crooke’s dark space, cathode glow
0.05	column splits into striations
0.01	cathode rays; the glass glows
10⁻⁴	no discharge

Gas	Air	H₂	N₂	Cl₂	CO₂	Neon
Colour	purple red	blue	red	green	bluish white	dark red

CATHODE RAYS

25.1.2 Properties of cathode rays

Streams of fast electrons, produced at about 10⁻² mm of Hg.
They travel in straight lines and leave the cathode normally.
They exert force, produce heat and cause fluorescence.
They are deflected by electric and magnetic fields.
They ionise gases and pass through thin metal foils.
Speed: from 130 to 110 of the speed of light.
On a heavy metal target they produce X-rays.

EMISSION

25.1.3 Ways of getting electrons out

Thermionic: by heating; thermions are electrons.
Photoelectric: by light. Secondary: by fast electrons hitting a surface.
An oxide coating lowers the work function, so a coated filament emits at a lower temperature.
X-ray emission gives out photons, not electrons.

THOMSON

25.1.4 Thomson’s experiment

Crossed E and B act on the beam.
Fields are adjusted so that the beam reaches P with no deflection.

Crossed fields act on the electron beam

eE = evB ⇒ v = EB

Anode voltage gives the speed.

eV = ½mv² ⇒ v = √(2eVm)

em = v²2V = E²2VB²

em = 1.77 × 10¹¹ Ckg, far above that of the proton.
It does not depend on the gas in the tube.
At large speed the mass grows, m = m₀ ÷ √(1 − v²c²), so em falls.
Switch off E, keep B: the beam moves in a circle.

DEFLECTION

25.1.5 Deflection in an electric field

The path is a parabola inside the plates

y = ½(eEm)l²v²

MILLIKAN

25.1.6 Millikan’s oil drop experiment

A balanced drop stays at rest

balanced drop: qE = mg, E = Vd

Charge on a drop is always a whole number times e.

q = ne, e = 1.6 × 10⁻¹⁹ C

JEE · NEET

Corner

Millikan: charge on the drop

q = 6πη(v₁ + v₂)dV [9ηv₁2g(ρ − σ)]^½

η = viscosity of air, d = gap between the plates, V = voltage.
v₁ = terminal speed with no field; v₂ = terminal speed with the field on.
ρ = density of oil, σ = density of air.

Positive rays

Positive ions pass through the holes in the cathode

Made at about 10⁻³ mm of Hg when fast electrons ionise gas atoms.
They carry a range of speeds, much smaller than cathode rays.
qm depends on the gas in the tube; it is largest for hydrogen.
Deflection in E and B is small compared with cathode rays.
Their charge is an integral multiple of e.
They cause more ionisation than cathode rays.
Their discovery helped in discovering isotopes.

Thomson’s mass spectrograph

E and B are parallel to each other, and both perpendicular to v.
Electric deflection is along y; magnetic deflection is along z.

y = qELDmv² · z = qBLDmv

z² = k(qm)y, k = B²LDE

One parabola for each value of qm

Ions of the same qm lie on one parabola; faster ions lie nearer the origin.
The number of parabolas gives the number of isotopes.

Bainbridge mass spectrograph

Velocity selector followed by the analysing field

Velocity selector (E ⟂ B) lets through only v = EB.
In the analysing field B′ the ions move on circles.

r = mEqBB′ · r₁r₂ = m₁m₂

separation of traces = 2v(m₂ − m₁)qB′

Practise PYQs on this topicCathode Rays, Thomson's Experiment & Gas Discharge · 12 questions with solutions

CONCEPT

Photons

In one line
Light travels as packets of energy called photons; each has energy hν, and its number decides the power.

ENERGY

25.2.1 Energy and momentum of a photon

E = hν = hcλ

E (in eV) = 12400λ (in Å) = 1240λ (in nm)

h = 6.63 × 10⁻³⁴ J s; 1 eV = 1.6 × 10⁻¹⁹ J.
Moves at speed c in vacuum.
Rest mass is zero; it is uncharged, so E and B do not deflect it.

p = Ec = hνc = hλ

effective mass m = Ec² = hcλ

Equal momentum ⇒ equal wavelength ⇒ equal energy.
Entering glass: frequency and energy stay the same; speed and wavelength fall.

NUMBER

25.2.2 Number of photons per second

n = PE = Phν = Pλhc

For the same power, longer wavelength ⇒ more photons per second.
If only a fraction η of the power becomes light, use ηP in place of P.
Monochromatic source in time t: number = nt.

INTENSITY

25.2.3 Intensity

Energy crossing unit area normally per second (W/m²).

I = EAt = PA

point source: I = P4πr² ⇒ I ∝ 1r²

PRESSURE

25.2.4 Radiation pressure and force

A reflected photon changes its momentum twice as much

Surface	Pressure	Force	Momentum given
absorbing	Ic	Pc	Uc
perfect reflector	2Ic	2Pc	2Uc
fraction r reflected	(1 + r)Ic	(1 + r)Pc	(1 + r)Uc

U = total energy falling on the surface; P = power.
Absorbs a part a, reflects a part r: F = Pc(a + 2r).

DUAL NATURE

25.2.5 Wave and particle

Light as a wave	Light as a particle
interference, diffraction, polarisation	photoelectric effect, Compton effect

Photoelectric effect cannot be explained by the wave theory.
One experiment shows only one nature at a time.

Practise PYQs on this topicPhotons: Energy, Momentum & Radiation Pressure · 49 questions with solutions

CONCEPT

Work Function & Einstein’s Equation

In one line
One photon gives all its energy to one electron: part pulls the electron out, the rest becomes its kinetic energy.

PHOTOELECTRIC

25.3.1 Photoelectric effect

Electrons (photoelectrons) leave a metal when light of suitable frequency falls on it.
Nearly all metals respond to UV; alkali metals (Li, Na, K, Rb, Cs) respond to visible light also.
Emission is instant: there is no time lag.

WORK FUNCTION

25.3.2 Work function, threshold frequency and wavelength

Work function W₀: minimum energy needed to remove an electron from the metal.
Threshold frequency ν₀: minimum frequency for emission.
Threshold wavelength λ₀: maximum wavelength for emission.

W₀ = hν₀ = hcλ₀

W₀ (in eV) = 12400λ₀ (in Å) = 1240λ₀ (in nm)

ν < ν₀ or λ > λ₀: no emission, however bright the light.
Larger W₀ ⇒ larger ν₀ ⇒ smaller λ₀.
For two metals: λ₀ ∝ 1W₀.

Metal	W₀ (eV)	Metal	W₀ (eV)
platinum	6.4	aluminium	4.3
gold	5.1	sodium	2.7
copper	4.7	lithium	2.5
silver	4.3	potassium	2.2
nickel	5.1	cesium	1.9

EINSTEIN

25.3.3 Einstein’s photoelectric equation

Photon at ν₀ just frees the electron; a larger hν gives it KE

hν = W₀ + K_max

K_max = hν − W₀ = h(ν − ν₀)

K_max = hc(1λ − 1λ₀)

K_max is for electrons at the surface.
Electrons from deeper inside lose energy on the way, so emitted electrons have a range of KE from 0 to K_max.

SPEED

25.3.4 Maximum speed and stopping potential

K_max = ½mv²_max = eV₀

v_max = √(2h(ν − ν₀)m)

V₀ is the stopping (retarding) potential, explained in the next concept.
K_max in eV has the same number as V₀ in volts: V₀ = 10 V ⇒ K_max = 10 eV.

Practise PYQs on this topicPhotoelectric Effect: Einstein's Equation · 175 questions with solutions

CONCEPT

Stopping Potential, Maximum Speed & KE Ratios

In one line
The negative plate voltage that just stops the fastest electron is the stopping potential: eV₀ = K_max.

SETUP

25.4.1 Experimental setup

Light falls on P; electrons travel to Q

P (cathode) and Q (anode) sit in an evacuated tube.
Light on P produces a photoelectric current in the outer circuit.
Q positive: current rises until it saturates.
Q negative: electrons are pushed back and current falls.
At Q = −V₀ even the fastest electron is stopped: current = 0.

K_max = eV₀

FORMULAS

25.4.2 Stopping potential formulas

V₀ = he(ν − ν₀)

V₀ = hce(1λ − 1λ₀)

V₀ (in volt) = 12400λ (in Å) − W₀ (in eV)

v_max = √(2hc(λ₀ − λ)mλλ₀)

TWO LIGHTS

25.4.3 Same metal, two lights

Write Einstein’s equation twice. W₀ is the same.

eV₁ = hcλ₁ − W₀, eV₂ = hcλ₂ − W₀

Subtract to remove W₀:

e(V₁ − V₂) = hc(1λ₁ − 1λ₂)

In terms of frequency:

h = e(V₁ − V₂)ν₁ − ν₂

RATIOS

25.4.4 Ratios of KE and speed

K₁K₂ = v₁²v₂² = V₁V₂ = hν₁ − W₀hν₂ − W₀

K_max is not proportional to ν: doubling ν more than doubles K_max.
v_max ∝ √(ν − ν₀), so it is not proportional to ν either.
Photon energy written in terms of W₀: hν = 3W₀ ⇒ K_max = 2W₀.
With ν = nν₀: K_max = (n − 1)W₀.

Photon energy	K_max	V₀
2W₀	W₀	W₀e
3W₀	2W₀	2W₀e
5W₀	4W₀	4W₀e

Photon from a hydrogen jump: n = 3 to n = 1 gives 12.09 eV; then use K_max = hν − W₀.

Practise PYQs on this topicPhotoelectric Experiment: Intensity, Stopping Potential & Graphs · 87 questions with solutions

CONCEPT

Effect of Intensity & Frequency

In one line
Intensity controls how many electrons come out; frequency controls how fast they come out.

INTENSITY

25.5.1 Effect of intensity (frequency fixed)

More intensity ⇒ more photons ⇒ more electrons per second.
Saturation current increases.
Stopping potential and K_max stay the same.

intensity ∝ photons per second ∝ electrons per second ∝ photocurrent

Same −V₀; saturation current ∝ intensity

FREQUENCY

25.5.2 Effect of frequency (intensity fixed)

Higher frequency ⇒ larger V₀ and larger K_max.
Saturation current is taken the same when the intensity is the same.

Same saturation; larger ν gives larger stopping potential

SUMMARY

25.5.3 What depends on what

Quantity	Intensity	Frequency	Metal
photocurrent	yes	no	no
K_max	no	yes	yes
V₀	no	yes	yes
ν₀, λ₀, W₀	no	no	yes

K_max and V₀ also do not depend on the accelerating voltage or on time.
Accelerating voltage changes the current only until saturation.

BELOW THRESHOLD

25.5.4 Below the threshold

Always check the frequency first.
New ν < ν₀ ⇒ photocurrent is zero, even if the intensity is doubled.
At ν = ν₀ electrons just get free with zero KE: taken as no current.
Example: 1.5ν₀ halved gives 0.75ν₀, so current is zero.

WAVE THEORY

25.5.5 Why the wave theory fails

Wave theory says K should grow with intensity: it does not.
Wave theory says any frequency should work if bright enough: there is a threshold.
Wave theory predicts a time lag: emission is instant.
Only the photon picture explains all three.

POINT SOURCE

25.5.6 Moving the source

I ∝ 1d² ⇒ current ∝ 1d²

Move a point source away: current falls.
V₀ and K_max do not change.

Practise PYQs on this topicPhotoelectric Experiment: Intensity, Stopping Potential & Graphs · 87 questions with solutions

CONCEPT

Graphs & Photocurrent

In one line
Every photoelectric graph comes from one equation: eV₀ = hν − W₀. Slope gives h; the intercept gives W₀.

V₀–ν

25.6.1 Stopping potential against frequency

V₀ = (he)ν − W₀e

Slope is he; it cuts the axis at ν₀

Slope = he; slope × e = h.
X-intercept = ν₀.
Y-intercept = −W₀e; so W₀ = e × (intercept length).

TWO POINTS

25.6.2 Planck’s constant from two points

h = e(V₂ − V₁)ν₂ − ν₁

KE–ν

25.6.3 Maximum KE against frequency

K_max = hν − W₀

Same shape; the slope is now h

Slope = h, for every metal.
X-intercept = ν₀; Y-intercept = −W₀.
The slope does not depend on the metal or on the intensity.

TWO METALS

25.6.4 Two different metals

Parallel lines: same slope, different ν₀

Lines are parallel for all metals.
The line further to the right has the larger ν₀, so the larger work function.
The ratio of slopes for any two metals is 1 : 1.

i–V

25.6.5 Reading current against voltage

Where the curve meets the negative axis: −V₀, which tells the frequency.
Height of the flat part: saturation current, which tells the intensity.
Same foot ⇒ same frequency. Same height ⇒ same intensity.

Current against intensity: straight line through O. V₀ against intensity: flat line.

NUMERICALS

25.6.6 Photocurrent numericals

photons per second on area A: n = IAhν

photocurrent i = n_e e, n_e = ηn

η = fraction of photons that free an electron (quantum efficiency).
Point source at d: I ∝ 1d², so the electrons per second change by the same factor.

Practise PYQs on this topicPhotoelectric Experiment: Intensity, Stopping Potential & Graphs · 87 questions with solutions

CONCEPT

de Broglie Wavelength: Momentum, KE & Voltage

In one line
Every moving particle has a wave with it, of wavelength λ = hp.

FORMULA

25.7.1 de Broglie wavelength

λ = hp = hmv = h√(2mE)

E = kinetic energy of the particle.
λ ∝ 1p ∝ 1v ∝ 1√E.

Same	λ then depends on
momentum	same λ for all
speed	λ ∝ 1m
kinetic energy	λ ∝ 1√m
voltage	λ ∝ 1√(mq)

CHANGES

25.7.2 KE changes

KE × 4 ⇒ λ becomes half.
KE × 16 ⇒ λ becomes one fourth: a 75% decrease.
To make λ half, KE must become 4 times, so add 3 times the old KE.

λ₁λ₂ = √(E₂E₁)

VOLTAGE

25.7.3 Charged particle through a voltage

E = ½mv² = qV

λ = h√(2mqV)

Particle	λ (in Å)
electron	12.27√V
proton	0.286√V
deuteron	0.202√V
α-particle	0.101√V

V in volts. Same particle: λ₁λ₂ = √(V₂V₁).

GRAPH

25.7.4 λ against voltage

Straight lines through the origin

Slope = h ÷ √(2mq).
For the same charge: smaller slope ⇒ larger mass.

COMPARE

25.7.5 Comparing particles

α-particle: m = 4m_p, q = 2e.

same V: λ_pλ_α = √(m_αq_αm_pq_p) = 2√2

same λ: V_αV_p = m_pq_pm_αq_α = 18

NATURE

25.7.6 Nature of matter waves

Tell the probability of finding the particle.
Not electromagnetic and not mechanical.
Exist for charged and uncharged moving particles alike.

Practise PYQs on this topicMatter Waves: de Broglie Wavelength · 151 questions with solutions

CONCEPT

Thermal Particles; Photon against Electron

In one line
Temperature gives a particle its KE; a photon and an electron with the same λ have the same momentum but different energy.

THERMAL

25.8.1 Thermal neutrons and gas molecules

Thermal neutron: E = kT.
Gas molecule at rms speed: E = (32)kT.
k = 1.38 × 10⁻²³ JK; T in kelvin.

neutron: λ = h√(2mkT) = 30.83√T Å

gas molecule: λ = h√(3mkT)

λ₁λ₂ = √(m₂T₂m₁T₁)

Change °C to kelvin first.
Same particle, T × 4 ⇒ λ becomes half: 27°C = 300 K to 927°C = 1200 K.

TWO FORMULAS

25.8.2 Photon and electron formulas

	Photon	Electron
λ from energy	hcE	h√(2mK)
energy from λ	hcλ	h²2mλ²
momentum	Ec	√(2mK)

SAME ENERGY

25.8.3 Same energy E

λ_phλ_e = cE√(2mE) = √(2mc²E)

λ_eλ_ph ∝ √E

p_php_e = √(E2mc²)

SAME WAVELENGTH

25.8.4 Same wavelength λ

They have the same momentum, not the same energy.

E_phK_e = 2mcλh = 2cv

If λ is the same and speed v of the electron is much less than c, the photon has much more energy.

LIGHT TO ELECTRON

25.8.5 Photoelectron’s wavelength

Light of wavelength λ falls on a metal; find λ_e of the fastest electron.
Negligible work function: K = hcλ.

λ_e = h√(2mK) = √(hλ2mc)

So λ_e ∝ √λ.
With a work function: find K first, then use the same formula.

K = hc(1λ − 1λ₀)

Practise PYQs on this topicMatter Waves: de Broglie Wavelength · 151 questions with solutions

CONCEPT

Fields, Bohr Orbits & Collisions

In one line
Find the momentum first in any situation, then write λ = hp.

E FIELD

25.9.1 Electron in an electric field

Force eE acts for time t, so momentum changes by eEt.
Starting from rest: p = eEt.

λ = heEt ⇒ dλdt = −heEt²

Force along v₀ (the electron speeds up):

λ = λ₀1 + (eEt ÷ mv₀)

Field perpendicular to v₀: speed grows sideways.

p² = (mv₀)² + (eEt)²

λ = λ₀√(1 + (eEt ÷ mv₀)²)

B FIELD

25.9.2 Charge in a magnetic field

Magnetic force does no work: speed stays the same.
So λ does not change, whichever way it is going.
On a circle of radius r: p = qBr.

λ = hqBr

BOHR

25.9.3 Bohr orbits

A stable orbit holds a whole number of waves.

Standing wave along the orbit

2πr = nλ

r ∝ n² and v ∝ 1n, so λ ∝ n.
First orbit: λ equals the circumference of the orbit, 3.3 Å.

λ_n = 2πa₀n ≈ 3.32n Å

Second excited state means n = 3; third excited state means n = 4.
In a magnetic-field orbit: 2πr = nhqBr, so r ∝ √n.

COLLISIONS

25.9.4 Decays and collisions

Momentum is conserved, so use it first.
Particle at rest breaks into two: momenta are equal and opposite, so λ is equal, whatever the masses.

Perfectly inelastic	Result
same direction	1λ = 1λ₁ + 1λ₂
opposite direction	1λ = \|1λ₁ − 1λ₂\|
at right angles	1λ² = 1λ₁² + 1λ₂²

Elastic head-on, B at rest: v_A = (m_A − m_B)v ÷ (m_A + m_B), v_B = 2m_Av ÷ (m_A + m_B).
Equal masses in the centre of mass frame: each has p = ½|p₁ − p₂|.

Practise PYQs on this topicOther / Mixed: Atoms & Bohr Links · 11 questions with solutions

CONCEPT

Davisson-Germer, Uncertainty, Compton & X-rays

In one line
Electrons diffract like waves, and photons collide like particles. Both natures were shown by experiment.

DAVISSON-GERMER

25.10.1 Davisson and Germer experiment

Electron beam falls on a nickel crystal

Verified the wave nature of electrons.
Nickel crystal acts as a diffraction grating.
Current in the filament changes the number of electrons.
Voltage between anode and filament changes their speed.

Scattered intensity has a peak at 50° for 54 V

BRAGG

25.10.2 Checking the wavelength

Bragg planes inside the crystal

2d sin θ = nλ

θ = 180° − φ2

d = 0.91 Å and θ = 65° give λ = 1.65 Å.
de Broglie gives 12.27 ÷ √54 = 1.67 Å: close agreement.
Using the spacing D between surface atoms: D sin φ = λ.

MICROSCOPE

25.10.3 Electron microscope

Resolution is of the order of λ of the electrons.
Electrons have far smaller λ than light, so better resolving power.
Same speed: λ ∝ 1m, so a proton microscope has about 1837 times the resolving power.

UNCERTAINTY

25.10.4 Heisenberg’s uncertainty principle

Position and momentum cannot both be known exactly at once.

Δx Δp ≥ h4π

ΔE Δt ≥ h4π

It explains why no electron exists in the nucleus.
Electrons through a narrow slit spread out like waves: angular width ∝ λa; a greater speed gives a smaller λ and narrower spread.

COMPTON

25.10.5 Compton effect

Photon scattered from an electron at rest

Δλ = λ′ − λ = hm₀c(1 − cos φ)

Scattered photon has less energy and a longer wavelength.
Energy and momentum are conserved.
Δλ = 0.0024 nm at φ = 90°; 0.0048 nm at φ = 180°.
Photon momentum hλ: use components to find the recoil angle.

X-RAYS

25.10.6 X-ray cut-off wavelength

An electron through V loses all its KE in one collision: this gives the most energetic photon.

eV = hcλ_min ⇒ λ_min = 12400V Å

λ_min depends only on V, not on the target.

electron of wavelength λ: λ_min = 2mcλ²h

JEE · NEET

Corner

Coolidge X-ray tube

Electrons from the hot filament hit the target

Filament current controls the intensity (number of electrons).
Voltage controls the quality (penetrating power).
Target: high atomic weight and high melting point (tungsten, molybdenum), set at 45°.
About 98% of the energy becomes heat; only about 2% becomes X-rays.

Properties of X-rays

Electromagnetic waves, λ from 0.1 Å to 100 Å.
Uncharged, so not deflected by E or B; travel at the speed of light.
λ_γ < λ_X < λ_UV.
Ionise gases, affect photographic plates, show diffraction and polarisation.
Lead is the best absorber; hydrogen emits no X-rays.

	Hard X-rays	Soft X-rays
penetration	more	less
frequency	about 10¹⁹ Hz	about 10¹⁶ Hz
wavelength	0.1 Å to 4 Å	4 Å to 100 Å

Absorption

I = I₀ e^−μx

half value thickness x_½ = 0.693μ

μ ∝ λ³ ∝ Z⁴.

Continuous X-rays

Higher voltage: smaller λ_min and greater intensity

A fast electron slows near a nucleus and gives out a photon (Bremsstrahlung).
All wavelengths above λ_min are present.

Characteristic X-rays

K series ends on n = 1; L series ends on n = 2

A fast electron knocks out an inner electron; an outer electron drops into the gap.
The sharp lines depend on the target (Z), not on the voltage.
λ_Kα < λ_Lα < λ_Mα and λ_Kα > λ_Kβ > λ_Kγ.

Sharp peaks on top of the continuous spectrum

Moseley’s law

√ν = a(Z − b)

Straight line for each series

Screening constant: b = 1 (K series), 7.4 (L series).
It showed that atomic number, not atomic weight, orders the periodic table.

1λ = R(Z − b)²(1n₁² − 1n₂²)

Kα: ν = 2.47 × 10¹⁵(Z − 1)² Hz

λ_Kα = 1216(Z − 1)² Å · E_Kα = 10.2(Z − 1)² eV

Practise PYQs on this topicDavisson-Germer, Uncertainty, Compton & X-Rays · 24 questions with solutions

EXAM-DAY CARD

Just see this before the exam

E = hν = hcλ

photon

W₀ = hν₀

work function

K_max = hν − W₀

Einstein

eV₀ = K_max

stopping potential

λ = hp

de Broglie

λ = 12.27√V Å

electron

FORMULA SHEET

Whole chapter on one page

Cathode rays

v = EB

em = E²2VB²

eV = ½mv²

q = ne

Photon

E (eV) = 12400λ (Å)

p = hλ = Ec

n = Pλhc

absorb: Ic; reflect: 2Ic

Photoelectric

W₀ = hν₀ = hcλ₀

K_max = h(ν − ν₀) = eV₀

V₀ = hce(1λ − 1λ₀)

h = e(V₁ − V₂)ν₁ − ν₂

Graphs

V₀–ν: slope he

K–ν: slope h

x-intercept: ν₀

i ∝ I; V₀ free of I

Matter waves

λ = h√(2mE) = h√(2mqV)

electron: 12.27√V Å

neutron: 30.83√T Å

gas: h√(3mkT)

Other

orbit: 2πr = nλ

Δx Δp ≥ h4π

Δλ = hm₀c(1 − cos φ)

λ_min = 12400V Å

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

More intensity gives larger K_max

K_max ∝ ν, so double ν gives double K

Using λ = hcE for an electron

Doubling intensity below ν₀ gives current

CORRECT ✓

Intensity changes the current only

K_max = hν − W₀: more than double

Electron: λ = h√(2mE)

Below ν₀ the current is always zero