JEE Main chapters · Class 12 · Physics

Dual Nature of Matter and Radiation: JEE Main Physics formulas and notes

Dual Nature of Matter and Radiation gives 1 or 2 questions in JEE Mains, almost every year. They are short numerical questions: the energy and number of photons, the stopping potential from Einstein’s equation, a graph with its slope and intercept, a comparison of two particles with the same K or the same V, and the Davisson–Germer value. Learn the five formulas by heart and practise the graphs; about 6 hours spread over 4 days is enough.

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Quick revision

• E = hf = hc/λ = 1240/λ(nm) eV; p = h/λ = E/c; N = P/hf; force of light P/c (absorbed), 2P/c (reflected)
• Photoemission needs f > f₀ (E > φ); no time lag; Kmax independent of intensity; current ∝ intensity
• Kmax = hf − φ = h(f − f₀); eV₀ = Kmax; V₀ (volt) = E (eV) − φ (eV)
• f doubled: K′ = 2K + φ; two frequencies: subtract the equations; two metals: K₁ − K₂ = φ₂ − φ₁
• K–f and V₀–f graphs: slope h or h/e (same for all metals), f-intercept f₀, other intercept −φ or −φ/e
• I–V curve: same f gives the same V₀; plateau Is ∝ intensity (photon rate); larger f gives a larger V₀
• λ = h/p = h/√(2mK) = h/√(2mqV); electron 12.27/√V Å, proton 0.286/√V Å, α 0.101/√V Å
• Same K: λ ∝ 1/√m; same V: λ ∝ 1/√(mq); same p: same λ; photon against electron: Eph/Ke = 2c/v
• Davisson–Germer 54 V, 50° peak, 2d sinθ = λ, λ = 0.165 nm; Bohr: 2πr = nλ; X-ray λmin = 12400/V Å

Formulas with solved examples

Photons and light quanta

Energy of a photon

• E = hf = (hc)/(λ)
• E in eV = (1240)/(λ in nm) = (12400)/(λ in Å)
• h = 6.63 × 10⁻³⁴ J s
• hc = 1240 eV nm = 1.99 × 10⁻²⁵ J m
• 1 eV = 1.6 × 10⁻¹⁹ J
• visible light, 400 nm to 700 nm, carries 3.1 eV to 1.8 eV
• a shorter wavelength means a more energetic photon

How to apply. Convert the wavelength to nm and divide 1240 by it to get the energy in eV at once. Keep J for h f and eV for comparing with the work function.

Example. A photon of 620 nm has E = 1240 ÷ 620 = 2.0 eV = 3.2 × 10⁻¹⁹ J. Violet light of 400 nm has 3.1 eV.

Momentum and mass of a photon

• p = (h)/(λ) = (E)/(c) = (hf)/(c)
• E = pc
• rest mass = 0, speed always c
• equivalent mass m = (E)/(c²) = (h)/(λc)
• a photon has no charge: it is not deflected by electric or magnetic fields

How to apply. Use p = h/λ for collisions and for the force of light. The equivalent mass is useful only to compare with other masses; it is not a rest mass.

Example. Photon of 500 nm: p = 6.63 × 10⁻³⁴ ÷ 5 × 10⁻⁷ = 1.33 × 10⁻²⁷ kg m/s and the equivalent mass is p/c = 4.4 × 10⁻³⁶ kg.

Number of photons and intensity

• N = (P)/(hf) = (Pλ)/(hc) photons per second from a source of power P
• intensity I = (P)/(A) = (photons per unit area per second) × hf
• point source: I = (P)/(4πr²), so I ∝ (1)/(r²)
• for 1 W at 500 nm, N = 2.5 × 10¹⁸ per second
• for the same power, a longer wavelength gives more photons

How to apply. Always write N = P/(hf) first. If only a fraction of the power is light, multiply P by that fraction (a lamp that is 5 % efficient gives 5 % of P as light).

Example. 1 W of light of 500 nm: N = 1 × 5 × 10⁻⁷ ÷ 1.99 × 10⁻²⁵ = 2.5 × 10¹⁸ photons per second.

Force and pressure of light

• light absorbed normally: F = (P)/(c) = (IA)/(c), pressure = (I)/(c)
• light reflected normally: F = (2P)/(c), pressure = (2I)/(c)
• fraction r reflected: F = (P(1 + r))/(c)
• energy density of the beam u = (I)/(c)
• the force comes from the change of the photon momentum

How to apply. Find the change of the momentum per second: p per photon times N for an absorbing surface, 2p per photon for a perfect mirror at normal incidence.

Example. A 3 W beam absorbed by a black plate pushes with F = 3 ÷ 3 × 10⁸ = 1 × 10⁻⁸ N. A perfect mirror normal to the beam gets 2 × 10⁻⁸ N.

Photoelectric effect: observations

Observations of Hertz and Lenard

• emission happens only if f ≥ f₀ (the threshold frequency of the metal), however strong the light
• for f > f₀ the number of photoelectrons per second (the current) ∝ intensity
• Kmax and V₀ depend on f and the metal, not on intensity
• emission is instantaneous (about 10⁻⁹ s) even for very faint light
• the classical wave theory fails on three counts: the threshold, Kmax independent of intensity, and the absence of any time lag
• Einstein (1905) explained it with photons, Millikan (1916) checked the equation and measured h

How to apply. Answer true or false questions with these five facts: threshold, current ∝ intensity, Kmax independent of intensity, no delay, Kmax rises linearly with f.

Example. Violet light is made twice as bright: twice as many electrons leave, with the same Kmax. Red light below f₀ however bright: no emission.

Work function and threshold

• φ = hf₀ = (hc)/(λ₀)
• f₀ = (φ)/(h)
• λ₀ = (1240)/(φ in eV) nm
• emission when f > f₀ or λ < λ₀
• φ depends only on the metal (Cs 2.14 eV, Na 2.75 eV, Cu 4.65 eV, Pt 5.65 eV)
• visible light can eject electrons only from the alkali metals (small φ)
• ultraviolet is needed for Cu, Zn, Pt

How to apply. Compare the photon energy with φ first: if E < φ nothing happens, whatever the intensity or the time.

Example. φ = 2.48 eV: λ₀ = 1240 ÷ 2.48 = 500 nm and f₀ = 2.48 × 1.6 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ = 6.0 × 10¹⁴ Hz.

Energy spread of the photoelectrons

• the emitted electrons have kinetic energies from 0 up to Kmax
• Kmax belongs to the electrons at the surface; an electron from deeper in the metal loses energy before it escapes
• the current falls smoothly to zero as the retarding voltage is raised to V₀
• the speed of the fastest electron vmax = √((2Kmax)/(m)) = 5.93 × 10⁵ √(V₀) m/s with V₀ in volt

How to apply. Whenever a question says maximum, use the surface electron and the stopping potential. An average speed is never asked.

Example. Kmax = 1.5 eV: electrons leave with 0 to 1.5 eV and a retarding potential of 1.5 V just stops all of them. vmax = 5.93 × 10⁵ × √1.5 = 7.3 × 10⁵ m/s.

Einstein’s equation and stopping potential

Einstein’s photoelectric equation

• hf = φ + Kmax
• Kmax = hf − φ = h(f − f₀) = hc((1)/(λ) − (1)/(λ₀))
• Kmax = (1)/(2)mvmax²
• one photon gives all its energy to one electron (all or nothing)
• this is the law of conservation of energy for one photon and one electron

How to apply. Write the photon energy in eV, subtract φ in eV, and you have Kmax in eV, which is also V₀ in volt. Only convert to J when you need the speed.

Example. λ = 4000 Å on a metal with φ = 2.0 eV: E = 3.1 eV, Kmax = 1.1 eV, vmax = 5.93 × 10⁵ × √1.1 = 6.2 × 10⁵ m/s.

Stopping potential

• eV₀ = Kmax
• V₀ = (h)/(e)(f − f₀) = (hc)/(e)((1)/(λ) − (1)/(λ₀))
• V₀ in volt = (photon energy in eV) − (φ in eV)
• V₀ is the reverse voltage at which the photocurrent just becomes zero
• V₀ does not depend on the intensity

How to apply. In numerical work V₀ in volt has the same number as Kmax in eV, so no charge needs to be put in.

Example. 300 nm on a metal with φ = 2.3 eV: E = 4.13 eV and V₀ = 4.13 − 2.3 = 1.83 V.

Two frequencies on the same metal

• (hf₁ − φ)/(hf₂ − φ) = (V₁)/(V₂) = (K₁)/(K₂)
• h = (e(V₁ − V₂))/(f₁ − f₂)
• φ = (e(V₁f₂ − V₂f₁))/(f₁ − f₂)
• the same metal means the same φ, so the difference of two Kmax equals the difference of the photon energies: K₁ − K₂ = h(f₁ − f₂)
• two metals in the same light: K₁ − K₂ = φ₂ − φ₁

How to apply. Write the equation twice and subtract to remove φ. Do not treat V₀ as proportional to f; it is proportional to f − f₀.

Example. For 4000 Å, V₁ = 1.1 V (so φ = 3.1 − 1.1 = 2.0 eV). For 3000 Å: eV₂ = 4.13 − 2.0 = 2.13 eV, so V₂ = 2.1 V.

Doubled frequency or halved wavelength

• if f is doubled, K′ = 2hf − φ = 2K + φ, which is more than 2K
• V₀′ = 2V₀ + (φ)/(e) > 2V₀
• if the wavelength is halved, the photon energy doubles, so the same result
• if the frequency is increased by a factor n, K′ = nK + (n − 1)φ

How to apply. A very common trap: Kmax does not double. Double the photon energy, then subtract φ once.

Example. K = 1 eV with φ = 2 eV: the photon energy is 3 eV. With f doubled: K′ = 6 − 2 = 4 eV (not 2 eV).

Electron reaching the collector

• accelerating voltage V on the collector: K at the collector = Kmax + eV
• retarding voltage V (V < V₀): K = Kmax − eV
• retarding voltage V ≥ V₀: no current
• the electrons arrive with every energy from (0 + eV) to (Kmax + eV) when accelerated

How to apply. Give the answer for the fastest electron: add or subtract eV from Kmax.

Example. Kmax = 2 eV. With +5 V on the collector the fastest electron arrives with 7 eV; with −1.5 V it has 0.5 eV; with −2 V none arrives.

Charging of an isolated plate or sphere

• an isolated metal body loses electrons and becomes positive until its potential equals V₀
• for a sphere of radius R: Q = 4πε₀RV₀ = (RV₀)/(9 × 10⁹)
• number of electrons lost N = (Q)/(e)
• the process stops when the potential is V₀ because the next electron cannot leave

How to apply. Use V = kQ/R with V = V₀ at the end. This is the only place where the capacitance of the body enters the chapter.

Example. Sphere of radius 10 cm, V₀ = 2 V: Q = 0.1 × 2 ÷ 9 × 10⁹ = 2.2 × 10⁻¹¹ C, that is 1.4 × 10⁸ electrons.

Graphs of the photoelectric effect

Kmax against frequency

• Kmax = hf − φ is a straight line
• slope = h (the same for every metal)
• intercept on the f axis = f₀
• intercept on the Kmax axis = −φ
• different metals give parallel lines; the metal with the smaller φ has its line to the left
• no point exists below f₀ (there is no emission)

How to apply. Compare slopes only to find h; compare the intercepts to compare φ. Parallel lines mean the same h, whatever the metal.

Example. A line cuts the f axis at 5 × 10¹⁴ Hz: φ = hf₀ = 6.63 × 10⁻³⁴ × 5 × 10¹⁴ = 3.3 × 10⁻¹⁹ J = 2.07 eV and the Kmax axis is cut at −2.07 eV.

Stopping potential against frequency

• V₀ = (h)/(e)f − (φ)/(e)
• slope = (h)/(e) = 4.14 × 10⁻¹⁵ V s (the same for every metal)
• intercept on the f axis = f₀
• intercept on the V₀ axis = −(φ)/(e)
• h = e × slope
• Millikan’s plot gave h = 6.57 × 10⁻³⁴ J s

How to apply. Read two points, find the slope, multiply by e. The intercept on the V₀ axis gives φ in eV directly as a negative number.

Example. A V₀–f line cuts the f axis at 5.0 × 10¹⁴ Hz and reaches 1.66 V at 9.0 × 10¹⁴ Hz: slope = 1.66 ÷ 4.0 × 10¹⁴ = 4.15 × 10⁻¹⁵ V s, h = 6.6 × 10⁻³⁴ J s and φ = 2.07 eV.

Photocurrent against voltage (I–V)

• the current rises with the collector voltage and saturates at Is
• Is ∝ the number of photons per second absorbed (the intensity at a fixed f)
• the curve meets the V axis at −V₀ on the retarding side
• same f, different intensity: same V₀, different Is
• different f, same photon rate: same Is, larger V₀ for larger f
• at equal power the higher frequency has fewer photons, so a smaller Is

How to apply. Read V₀ from the point on the negative side and Is from the plateau. A curve that starts further to the left has the larger frequency; a curve with a higher plateau has the larger intensity.

Example. Two curves share V₀ but the plateaus are in the ratio 1 : 2: same f, intensity ratio 1 : 2. Two curves with V₀ = 1 V and 2 V: the second light has the higher frequency.

Is against intensity and against frequency

• at a fixed f > f₀: Is is a straight line through the origin against intensity
• Is against f at constant power: zero up to f₀, then it falls like (1)/(f) (fewer photons)
• Is against f at a constant photon rate: a step at f₀, then flat
• V₀ against intensity: a horizontal line

How to apply. Decide first what is kept constant: power or the number of photons per second. The shape of the graph follows from it.

Example. Doubling the power of the same light doubles Is. The line Is against power passes through the origin and V₀ does not change.

Intensity, current and the photocell

Saturation current and efficiency

• Is = ηNe = η(P)/(hf)e = η(Pλe)/(hc)
• η = electrons emitted per photon absorbed (a few per thousand for most metals, often given as a percentage)
• the maximum possible current is for η = 1: Is = (Pλe)/(hc)
• Is ∝ P at a fixed f

How to apply. Find N per second first, then the electrons per second (ηN), then multiply by e.

Example. 1 mW of 400 nm light, η = 1 %: N = 1 × 10⁻³ × 4 × 10⁻⁷ ÷ 1.99 × 10⁻²⁵ = 2.0 × 10¹⁵ per second, so Is = 0.01 × 2.0 × 10¹⁵ × 1.6 × 10⁻¹⁹ = 3.2 μA.

Distance, slit and filter changes

• point source: I ∝ (1)/(r²) so Is ∝ (1)/(r²) and V₀ stays the same
• a slit or a screen that cuts the intensity cuts Is, not V₀
• a filter that changes the frequency changes V₀ (and the photon rate)
• a larger collector at the same voltage cannot raise Is beyond the saturation value

How to apply. Separate what depends on intensity (Is, the number of electrons) from what depends on frequency (V₀, Kmax, vmax).

Example. Lamp moved from 20 cm to 40 cm: the intensity falls to one fourth, so Is becomes one fourth; V₀ is unchanged.

Same light, two metals

• K₁ − K₂ = φ₂ − φ₁ for one beam and two metals
• V₁ − V₂ = (φ₂ − φ₁)/(e)
• the metal with the smaller φ has the larger Kmax and the larger V₀
• both stopping potentials move up together if the frequency is raised

How to apply. Subtract the two Einstein equations; hf cancels, so the answer does not depend on the light.

Example. φ₁ = 2.0 eV, φ₂ = 3.0 eV and a 4 eV photon: K₁ = 2.0 eV, K₂ = 1.0 eV and the difference is 1.0 eV = φ₂ − φ₁.

Matter waves and de Broglie wavelength

de Broglie wavelength

• λ = (h)/(p) = (h)/(mv)
• in terms of kinetic energy: λ = (h)/(√(2mK))
• accelerated through V: λ = (h)/(√(2mqV))
• every moving particle has a wave; the wave is noticeable only for tiny masses
• λ does not depend on the charge of the particle, only on p
• the matter wave is not an electromagnetic wave

How to apply. Pick the form that matches the data: v given, K given or V given. For V use q = e for an electron or a proton and 2e for an α particle.

Example. A ball of 0.1 kg at 10 m/s has λ = 6.63 × 10⁻³⁴ ÷ 1 = 6.6 × 10⁻³⁴ m, far too small to see. An electron at 100 V: λ = 12.27 ÷ 10 = 1.23 Å.

Ready numerical forms

• electron: λ = (12.27)/(√V) Å = (1.227)/(√V) nm (V in volt)
• proton: λ = (0.286)/(√V) Å
• α particle: λ = (0.101)/(√V) Å
• neutron of energy E in eV: λ = (0.286)/(√E) Å
• a thermal particle: λ = (h)/(√(3mkT))
• neutron at 300 K: about 1.45 Å
• these hold only for non-relativistic speeds

How to apply. Use them to save time and learn the three constants 12.27, 0.286 and 0.101. They come from the formula with the mass and the charge in the root.

Example. Proton at 100 V: 0.286 ÷ 10 = 0.0286 Å. Neutron of 0.04 eV: 0.286 ÷ 0.2 = 1.43 Å.

Comparing particles

• same K: λ ∝ (1)/(√m)
• same V: λ ∝ (1)/(√(mq))
• same speed: λ ∝ (1)/(m)
• same momentum: same λ
• proton against α particle: same V gives λp : λα = 2√2 : 1, same K gives 2 : 1, same speed gives 4 : 1
• electron against proton at the same K: λe : λp = √(mp/me) = 43

How to apply. Decide which quantity is common (K, V, v or p) and write λ in that quantity only.

Example. An electron and a proton have the same kinetic energy: λe ÷ λp = √1836 = 43, so the electron wave is much longer.

Photon and electron of equal wavelength

• for the same λ: Eph ÷ Ke = (2c)/(v) (photon energy pc, electron energy (p²)/(2m))
• for the same energy: λph ÷ λe = (2c)/(v) = c√((2m)/(E))
• the photon wavelength is much longer for the same energy
• momentum p = h/λ is the same in the first case

How to apply. Write the photon energy as pc and the electron energy as p²/2m, then divide.

Example. An electron of 100 eV has v = 5.9 × 10⁶ m/s. A photon with the same wavelength has 2c/v = 101 times the energy, 10.1 keV.

Percentage change of λ

• λ ∝ V^−1/2
• V increased by 21 %: λ falls to (1)/(1.1), a decrease of 9.1 %
• V made four times: λ halves
• small changes: (Δλ)/(λ) = −(1)/(2)(ΔV)/(V)
• the same rule holds with K in place of V

How to apply. Write λ′ = λ ÷ √(V′/V). Do not subtract percentages directly.

Example. An electron’s accelerating voltage is raised by 21 %: λ′ = λ ÷ 1.1 = 0.909λ, a fall of about 9 %.

Charged particle in a magnetic field

• radius r = (p)/(qB) gives p = qBr
• λ = (h)/(qBr)
• the same result is true for any charged particle moving on a circle in a uniform field

How to apply. The orbit radius gives the momentum at once; no speed is needed.

Example. An electron circling with r = 1 mm in B = 1 mT has p = 1.6 × 10⁻¹⁹ × 10⁻³ × 10⁻³ = 1.6 × 10⁻²⁵ kg m/s and λ = 4.1 nm.

Davisson–Germer, Bohr link and X-rays

Davisson–Germer experiment

• a beam of electrons accelerated through 54 V is scattered from a nickel crystal
• the intensity has a peak at a scattering angle of 50°
• Bragg: 2d sinθ = nλ with d = 0.091 nm, θ = 65° gives λ = 0.165 nm
• de Broglie: λ = 12.27 ÷ √54 = 0.167 nm, in agreement
• the scattering angle is φ = 180° − 2θ
• it proved the wave nature of electrons; G P Thomson saw rings with a thin foil

How to apply. Use the two values as a check: the formula and the Bragg law agree. In problems, apply 2d sinθ = nλ with θ the glancing angle and find λ or V.

Example. 54 V: λ = 12.27 ÷ 7.35 = 1.67 Å, and Bragg’s law with d = 0.91 Å at θ = 65° gives 2 × 0.91 × sin 65° = 1.65 Å.

Diffraction and the electron microscope

• diffraction needs λ about the same as the spacing of the scatterers (about 0.1 nm in a crystal)
• an electron of about 100 V has the right wavelength for a crystal
• the resolving limit of a microscope is about λ, and an electron beam of a few kV has a wavelength thousands of times smaller than visible light
• a higher accelerating voltage means a smaller λ and a sharper image

How to apply. Compare wavelengths: whichever is much smaller than the object shows detail; a wave much longer than the gaps bends around.

Example. Electrons at 10 kV: λ = 12.27 ÷ 100 = 0.123 Å, about 4000 times shorter than green light (5000 Å).

Bohr orbit as a standing wave

• the circumference holds a whole number of wavelengths: 2πr = nλ = n(h)/(mv)
• this gives mvr = n(h)/(2π)
• λn = (2πrn)/(n) = 3.32 n Å for hydrogen (rn = 0.529 n² Å)
• in the nth orbit the number of waves is n

How to apply. Replace mv by h/λ and the angular momentum rule follows at once. For the wavelength, divide the circumference by n.

Example. Hydrogen, n = 2: r = 4 × 0.529 = 2.12 Å and λ = 2π × 2.12 ÷ 2 = 6.6 Å.

X-rays: the cutoff wavelength

• electrons accelerated through V and stopped in a target produce X-rays
• the greatest photon energy is eV, so λmin = (hc)/(eV) = (12400)/(V in volt) Å
• λmin depends only on V, not on the target
• it is the photoelectric effect run backwards (Duane–Hunt law)

How to apply. The shortest wavelength belongs to the case when all the electron energy goes into one photon.

Example. V = 20 kV: λmin = 12400 ÷ 20000 = 0.62 Å.

Recoil on emitting or absorbing a photon

• momentum is conserved: the atom takes p = (h)/(λ) in the opposite direction
• recoil speed v = (h)/(mλ)
• recoil energy (h²)/(2mλ²), very small against the photon energy
• an atom that absorbs a photon moves with the same v

How to apply. Write conservation of momentum for the photon and the atom; the recoil energy is only needed in precision questions.

Example. A hydrogen atom emitting a 10.2 eV photon (121.6 nm): v = 6.63 × 10⁻³⁴ ÷ (1.67 × 10⁻²⁷ × 1.216 × 10⁻⁷) = 3.3 m/s.

Wave–particle duality in one line

• light shows interference and diffraction (waves) and the photoelectric effect (particles)
• electrons show tracks (particles) and diffraction (waves)
• the intensity at a point ∝ (amplitude)² ∝ the number of photons that arrive there
• a particle’s wave is a probability wave: |amplitude|² gives the chance of finding it

How to apply. Choose the model that fits the experiment: propagation and interference by waves, exchange of energy and momentum by particles.

Example. The bright fringes in a double-slit experiment are where many photons or electrons arrive one by one; the pattern builds up from single hits.

Most asked question types

How do I solve “Energy, momentum and mass of a photon” questions in JEE Main?

The wavelength or frequency of a photon is given. Find its energy in eV, its momentum or its equivalent mass, or compare two photons. 1. E = 1240 ÷ λ(nm) eV. 2. p = h/λ = E/c. 3. Equivalent mass m = E/c². Example: 500 nm: 2.48 eV, p = 1.33 × 10⁻²⁷ kg m/s.

How do I solve “Photons per second and the force of light” questions in JEE Main?

The power of a source and the wavelength are given. Find the number of photons per second, the intensity at a distance or the force on an absorbing or reflecting surface. 1. N = P/hf = Pλ/hc. 2. I = P/4πr² for a point source. 3. F = P/c (absorbed) or 2P/c (reflected). Example: 1 W, 500 nm: 2.5 × 10¹⁸ per second.

How do I solve “Threshold wavelength and work function” questions in JEE Main?

The work function in eV or the threshold wavelength is given. Find the other one, or say whether a given light will eject electrons. 1. λ₀ = 1240 ÷ φ(eV) nm. 2. Emission needs λ < λ₀. 3. f₀ = φ/h. Example: φ = 2.48 eV: λ₀ = 500 nm.

How do I solve “True or false about the photoelectric effect” questions in JEE Main?

Several statements about intensity, frequency, time lag, current and Kmax are given. Choose the correct ones. 1. Kmax depends on f and the metal. 2. The current depends on the intensity and the photon rate. 3. Emission is instantaneous and needs f > f₀. Example: Doubling the intensity doubles the current and leaves V₀ unchanged.

How do I solve “Maximum kinetic energy and stopping potential” questions in JEE Main?

The wavelength of light and the work function are given. Find Kmax, V₀ or the greatest speed. 1. Photon energy in eV minus φ is Kmax in eV. 2. V₀ in volt equals Kmax in eV. 3. vmax = 5.93 × 10⁵ √V₀ m/s. Example: 4000 Å, φ = 2 eV: Kmax = 1.1 eV.

How do I solve “Two wavelengths or two frequencies on the same metal” questions in JEE Main?

Stopping potentials are given for two lights. Find φ, h, f₀ or the stopping potential for a third light. 1. Write hf − φ = eV twice. 2. Subtract to cancel φ, or use the ratio. 3. Put the result in the third equation. Example: 1.1 V at 4000 Å gives φ = 2.0 eV and V₀ = 2.1 V at 3000 Å.

How do I solve “Frequency doubled or wavelength halved” questions in JEE Main?

Kmax or V₀ is given for a light and the frequency is changed by a factor. Find the new Kmax or V₀. 1. Photon energy scales with f, not Kmax. 2. K′ = nhf − φ. 3. Compare with 2K: K′ = 2K + φ. Example: K = 1 eV, φ = 2 eV: K′ = 4 eV when f is doubled.

How do I solve “Electron reaching the anode” questions in JEE Main?

An accelerating or a retarding voltage is applied. Find the greatest energy with which an electron arrives, or the voltage at which the current stops. 1. Accelerating: K = Kmax + eV. 2. Retarding: K = Kmax − eV. 3. The current stops when the retarding voltage is V₀. Example: Kmax = 2 eV, +5 V: 7 eV.

How do I solve “Slope and intercept of Kmax–f and V₀–f graphs” questions in JEE Main?

A straight-line graph is given. Find h, the threshold frequency, the work function, or tell what changes for another metal. 1. Slope = h (or h/e). 2. f intercept = f₀. 3. Other intercept = −φ (or −φ/e). Example: f intercept 5 × 10¹⁴ Hz: φ = 2.07 eV.

How do I solve “I–V curves for different intensity or frequency” questions in JEE Main?

Two or three photocurrent curves are drawn. Say which light has the larger intensity or frequency, or find V₀ and Is. 1. Same V₀: same frequency. 2. A higher plateau: more photons. 3. Further to the left: higher frequency. Example: Same V₀, plateau ratio 1 : 2: intensity ratio 1 : 2.

How do I solve “Distance, intensity and filters” questions in JEE Main?

The lamp is moved, the intensity is changed or a filter is used. Say what happens to the current and to the stopping potential. 1. I ∝ 1/r²; Is ∝ I. 2. V₀ does not depend on the intensity. 3. A filter that changes f changes V₀. Example: Distance doubled: Is becomes one fourth, V₀ is the same.

How do I solve “Photocurrent from the power and the efficiency” questions in JEE Main?

The power, the wavelength and the fraction of photons that eject an electron are given. Find the saturation current. 1. N = P/hf per second. 2. Electrons per second = ηN. 3. Is = ηNe. Example: 1 mW, 400 nm, η = 1 %: 3.2 μA.

How do I solve “Charging of an isolated sphere or plate” questions in JEE Main?

An isolated metal body is lit until it stops emitting. Find the charge or the number of electrons lost. 1. The potential becomes V₀. 2. Q = 4πε₀RV₀. 3. N = Q/e. Example: R = 10 cm, V₀ = 2 V: Q = 2.2 × 10⁻¹¹ C.

How do I solve “Wavelength of an electron from the voltage or energy” questions in JEE Main?

An electron is accelerated through V volt or has a given kinetic energy. Find λ, or the voltage that gives a stated λ. 1. λ = 12.27/√V Å. 2. λ = h/√(2mK) for a given K. 3. V = (12.27/λ)² with λ in Å. Example: 100 V: 1.23 Å.

How do I solve “Comparing electron, proton, neutron and α particle” questions in JEE Main?

Two particles are accelerated by the same voltage, or have the same energy, speed or momentum. Find the ratio of their wavelengths. 1. Fix the common quantity. 2. Same K: λ ∝ 1/√m. 3. Same V: λ ∝ 1/√(mq). Example: Proton against α at the same V: 2√2 : 1.

How do I solve “Photon and electron of the same wavelength or energy” questions in JEE Main?

A photon and an electron are compared at the same λ or at the same energy. Find the ratio of the energies or wavelengths. 1. Eph = pc, Ke = p²/2m. 2. Eph/Ke = 2c/v. 3. Same energy: λph/λe = 2c/v. Example: 100 eV electron: the photon has 10.1 keV.

How do I solve “Percentage change of the wavelength” questions in JEE Main?

The accelerating voltage or the energy is changed by a percentage. Find the percentage change of λ. 1. λ ∝ V^−1/2. 2. λ′ = λ ÷ √(V′/V). 3. Do not subtract percentages directly. Example: V up by 21 %: λ down by about 9 %.

How do I solve “Davisson–Germer and Bragg’s law” questions in JEE Main?

Electrons are scattered from a crystal. Find the wavelength, the spacing or the angle of the strongest peak. 1. λ = 12.27/√V. 2. 2d sinθ = nλ, θ the glancing angle. 3. Scattering angle φ = 180° − 2θ. Example: 54 V: 1.67 Å, 2 × 0.91 × sin 65° = 1.65 Å.

How do I solve “Bohr orbit as a standing wave” questions in JEE Main?

An electron is in the nth orbit of hydrogen. Find the de Broglie wavelength or the number of waves around the orbit. 1. 2πr = nλ. 2. λ = 2πr/n. 3. mvr = nh/2π follows. Example: n = 2: r = 2.12 Å, λ = 6.6 Å.

How do I solve “Cutoff wavelength of X-rays” questions in JEE Main?

Electrons are accelerated through a voltage and strike a target. Find the shortest wavelength of the X-rays. 1. The greatest photon energy is eV. 2. λmin = 12400/V Å. 3. It does not depend on the target. Example: 20 kV: 0.62 Å.

Common myths and the truth

• Myth: A very strong red light can eject electrons from a metal whose threshold is in the ultraviolet. Truth: Each electron absorbs one photon. If the photon energy is below φ nothing happens, however many photons arrive.
• Myth: Brighter light gives electrons with more kinetic energy. Truth: A brighter light gives more electrons. The Kmax depends only on the frequency and the metal.
• Myth: The photoelectric effect shows that light is only a particle. Truth: It needs the photon picture, but interference and diffraction still need the wave picture. Light has both properties.
• Myth: A photon has a mass, so it can be stopped. Truth: A photon has no rest mass and always moves at c. The equivalent mass E/c² describes only its energy.
• Myth: Doubling the frequency doubles the stopping potential. Truth: V₀ = (h/e)(f − f₀). Doubling f gives more than double V₀.
• Myth: Only electrons have a wavelength. Truth: Every moving particle has λ = h/p; a ball’s wave is too small to be noticed.
• Myth: A heavier particle always has a shorter wavelength. Truth: It depends on what is common. At the same speed λ ∝ 1/m; at the same momentum all have the same λ.
• Myth: The matter wave is an electromagnetic wave. Truth: It is a probability wave; it has no electric or magnetic field.
• Myth: The Davisson–Germer peak at 50° is the Bragg angle. Truth: 50° is the scattering angle. The Bragg glancing angle is 65°, since φ = 180° − 2θ.
• Myth: X-rays come out with a single energy. Truth: The spectrum is continuous down to λmin = 12400/V Å, with sharp lines of the target added to it.

Mistakes to avoid

• Kmax does not double when the frequency doubles: K′ = 2K + φ.
• V₀ is proportional to f − f₀, not to f.
• The stopping potential does not depend on the intensity; the saturation current does.
• The slope of the Kmax–f and V₀–f graph is the same for all metals; the intercepts differ.
• At equal power a higher frequency has fewer photons, so a smaller current, not a larger one.
• 12.27/√V Å is for an electron only; the proton and α forms have other constants.
• For the same V the α particle has q = 2e, and λ ∝ 1/√(mq).
• In the Davisson–Germer formula θ is the glancing angle, φ = 180° − 2θ is the scattering angle.

Simulations for Dual Nature of Matter and Radiation

• Compton scattering — Scatter an X-ray photon from a free electron and follow the sharing of energy and momentum.
• Matter-wave wavelength — Accelerate an electron through a potential difference and calculate its wavelength.
• Electron diffraction from a crystal — Change electron energy and locate the first-order Bragg reflection.
• Photocell current vs anode voltage — Vary the anode voltage of a photocell. Intensity sets the saturation current; frequency sets the stopping…
• Photoelectric effect — Shine light on a metal and see whether it releases electrons.
• Stopping potential vs frequency — Shine different frequencies on a metal and read the stopping voltage.

More: Class 12 Physics simulations · all JEE Main chapters.