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Modern Physics flash cards

Master Modern Physics through 94 JEE Advanced-level recall cards, systematically structured one idea at a time. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

Modern Physics, question and answer

21 of this chapter's 94 cards, laid out open so you can read straight through. The remaining 73 are in the interactive deck, where the answer stays hidden until you commit to one.

  1. 1.State Einstein's photoelectric equation.

    Kmax=hνϕ0K_{max}=h\nu-\phi_0, where ϕ0\phi_0 is the work function. Equivalently eV0=hνϕ0eV_0=h\nu-\phi_0, with V0V_0 the stopping potential.

    Hint: Photon energy = work function + max KE of ejected electron.

  2. 2.What is the work function ϕ0\phi_0 of a metal?

    The minimum energy needed to just liberate an electron from the metal surface. Related to threshold frequency by ϕ0=hν0\phi_0=h\nu_0 and to threshold wavelength by ϕ0=hc/λ0\phi_0=hc/\lambda_0.

    Hint: Energy to escape the surface, no leftover KE.

  3. 3.Define threshold frequency ν0\nu_0.

    The minimum frequency of incident light below which no photoemission occurs, regardless of intensity. ν0=ϕ0/h\nu_0=\phi_0/h.

    Hint: Below it, even bright light gives no electrons.

  4. 4.What is the stopping potential V0V_0?

    The reverse voltage that just stops the most energetic photoelectrons: eV0=Kmax=hνϕ0eV_0=K_{max}=h\nu-\phi_0. It depends on frequency, not intensity.

    Hint: Voltage that makes photocurrent zero.

  5. 5.How does photoelectric current depend on light intensity (at fixed frequency >ν0>\nu_0)?

    Saturation photocurrent is directly proportional to intensity (more photons → more electrons). Stopping potential and KmaxK_{max} are unaffected by intensity.

    Hint: Intensity sets number of electrons, not their energy.

  6. 6.How does KmaxK_{max} of photoelectrons depend on frequency?

    Linearly: Kmax=hνϕ0K_{max}=h\nu-\phi_0. A graph of KmaxK_{max} vs ν\nu is a straight line of slope hh and xx-intercept ν0\nu_0.

    Hint: Slope of the line is Planck's constant.

  7. 7.On a V0V_0 vs ν\nu graph, what do the slope and intercepts represent?

    Slope =h/e=h/e (same for all metals). xx-intercept =ν0=\nu_0 (threshold frequency). yy-intercept magnitude =ϕ0/e=\phi_0/e.

    Hint: Parallel lines for different metals; slope is universal.

  8. 8.Why does the photoelectric effect show no time lag?

    Emission is essentially instantaneous (<109<10^{-9} s) because a single photon delivers its whole energy to one electron at once, contradicting the wave prediction of gradual energy build-up.

    Hint: One-photon-one-electron, immediate energy transfer.

  9. 9.What is a photon and its energy/momentum?

    A quantum of light with energy E=hν=hc/λE=h\nu=hc/\lambda and momentum p=E/c=h/λp=E/c=h/\lambda. It has zero rest mass and travels at cc.

    Hint: Massless packet; p=h/λp=h/\lambda.

  10. 10.Give the convenient numerical form for photon energy in eV.

    E(eV)=1240λ(nm)E(\text{eV})=\dfrac{1240}{\lambda(\text{nm})}. So 620620 nm light 2\approx 2 eV.

    Hint: Remember 1240 eV·nm.

  11. 11.State the de Broglie wavelength of a matter wave.

    λ=hp=hmv\lambda=\dfrac{h}{p}=\dfrac{h}{mv}. Every moving particle has an associated wave; the effect is significant only for small mvmv.

    Hint: Wavelength = h over momentum.

  12. 12.Express de Broglie wavelength in terms of kinetic energy KK.

    λ=h2mK\lambda=\dfrac{h}{\sqrt{2mK}}. For a charge accelerated through potential VV: λ=h2mqV\lambda=\dfrac{h}{\sqrt{2mqV}}.

    Hint: p=2mKp=\sqrt{2mK}.

  13. 13.de Broglie wavelength of an electron accelerated through VV volts (numerical).

    λ=12.27V A˚\lambda=\dfrac{12.27}{\sqrt{V}}\ \text{Å} (i.e. 1.227\approx 1.227 nm/V/\sqrt{V}). For V=100V=100 V, λ1.23\lambda\approx1.23 Å.

    Hint: 12.27 Å over root V.

  14. 14.What experiment confirmed the wave nature of electrons?

    The Davisson–Germer experiment (1927): electrons scattered off a nickel crystal showed diffraction maxima matching the de Broglie wavelength.

    Hint: Electron diffraction off a crystal.

  15. 15.State the postulates of Bohr's model of hydrogen.

    (1) Electrons orbit in stationary states without radiating. (2) Angular momentum is quantized: mvr=nh2πmvr=n\dfrac{h}{2\pi}. (3) Radiation on jumps: hν=EiEfh\nu=E_i-E_f.

    Hint: Quantized L, stationary orbits, quantum jumps.

  16. 16.How does Bohr's quantization link to de Broglie?

    A standing wave fits the orbit: 2πr=nλ=nhmv2\pi r=n\lambda=n\dfrac{h}{mv}, giving mvr=nh2πmvr=n\dfrac{h}{2\pi}. Only whole numbers of wavelengths survive.

    Hint: Circumference = integer × wavelength.

  17. 17.Bohr radius formula for hydrogen-like atom.

    rn=n2Za0r_n=\dfrac{n^2}{Z}\,a_0, with a0=0.529a_0=0.529 Å =ε0h2πme2=\dfrac{\varepsilon_0 h^2}{\pi m e^2}. Radius grows as n2n^2, shrinks with ZZ.

    Hint: rn2/Zr\propto n^2/Z.

  18. 18.Bohr energy of the nn-th level (hydrogen-like).

    En=13.6Z2n2 eVE_n=-13.6\dfrac{Z^2}{n^2}\ \text{eV}. Negative (bound); magnitude falls as 1/n21/n^2.

    Hint: 13.6Z2/n2-13.6\,Z^2/n^2 eV.

  19. 19.Bohr orbital velocity of the electron.

    vn=Zn×2.19×106 m/s=αcZnv_n=\dfrac{Z}{n}\times 2.19\times10^6\ \text{m/s}=\alpha c\dfrac{Z}{n}, where α1/137\alpha\approx 1/137.

    Hint: vZ/nv\propto Z/n; ground state c/137\approx c/137.

  20. 20.Relations among KE, PE and total energy in Bohr atom.

    K=EK=-E, U=2EU=2E, so U=2KU=-2K and E=K=U/2E=-K=U/2. Total energy magnitude equals kinetic energy.

    Hint: Virial theorem: PE = −2·KE.

  21. 21.What is the ionization energy of hydrogen?

    13.613.6 eV — energy to remove the ground-state (n=1n=1) electron to n=n=\infty. For hydrogen-like: 13.6Z213.6\,Z^2 eV.

    Hint: Energy from n=1n=1 to free.

Open the interactive deck for the other 73 cards, with self-grading so the ones you keep missing come back.

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