How to Use This Problem Set

Your full workout for Dual Nature, grouped by theme: photon energetics, work function & threshold, Einstein's equation & stopping potential, graphs & slopes, and de Broglie wavelengths.

Keep these handy:

  • Photon: E=hν=hcλE = h\nu = \dfrac{hc}{\lambda}; p=hλp = \dfrac{h}{\lambda}; photons per second N=PhνN = \dfrac{P}{h\nu}
  • Handy constant: hc≈1240hc \approx 1240 eV nm; 1 eV = 1.6×10−191.6 \times 10^{-19} J
  • Threshold: ϕ0=hν0=hcλ0\phi_0 = h\nu_0 = \dfrac{hc}{\lambda_0}
  • Einstein: Kmax=hν−ϕ0=eV0K_{max} = h\nu - \phi_0 = eV_0; V0=heν−ϕ0eV_0 = \dfrac{h}{e}\nu - \dfrac{\phi_0}{e} (slope h/eh/e)
  • Max speed: vmax=2Kmax/mv_{max} = \sqrt{2K_{max}/m}
  • de Broglie: λ=hp=hmv=h2mK\lambda = \dfrac{h}{p} = \dfrac{h}{mv} = \dfrac{h}{\sqrt{2mK}}; electron through V volts: λ≈1.227V\lambda \approx \dfrac{1.227}{\sqrt{V}} nm

State the condition, then substitute.

Solved Examples - Photon Energetics

Example 1. Energy of a 620 nm photon in eV?

Solution: E=1240620=2.0E = \dfrac{1240}{620} = 2.0 eV.

Example 2. Frequency of a 3.0 eV photon?

Solution: ν=Eh=3.0×1.6×10−196.63×10−34≈7.2×1014\nu = \dfrac{E}{h} = \dfrac{3.0 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 7.2 \times 10^{14} Hz.

Example 3. Momentum of a 500 nm photon?

Solution: p=hλ=6.63×10−345×10−7=1.33×10−27p = \dfrac{h}{\lambda} = \dfrac{6.63 \times 10^{-34}}{5 \times 10^{-7}} = 1.33 \times 10^{-27} kg m/s.

Example 4. A 5 mW laser emits 550 nm light. Photons per second?

Solution: E=1240550=2.25E = \dfrac{1240}{550} = 2.25 eV =3.6×10−19= 3.6 \times 10^{-19} J; N=5×10−33.6×10−19≈1.4×1016N = \dfrac{5 \times 10^{-3}}{3.6 \times 10^{-19}} \approx 1.4 \times 10^{16} per second.

Example 5. A radio station broadcasts 10 kW at 100 MHz. Photons per second?

Solution: E=hν=(6.63×10−34)(108)=6.63×10−26E = h\nu = (6.63 \times 10^{-34})(10^8) = 6.63 \times 10^{-26} J; N=1046.63×10−26≈1.5×1029N = \dfrac{10^4}{6.63 \times 10^{-26}} \approx 1.5 \times 10^{29} per second — so many that graininess is undetectable.

Example 6. Energy of an X-ray photon of wavelength 0.1 nm?

Solution: E=12400.1=12400E = \dfrac{1240}{0.1} = 12400 eV = 12.4 keV.

Example 7. Ratio of energies of photons at 400 nm and 800 nm?

Solution: E∝1/λE \propto 1/\lambda: ratio =800/400=2:1= 800/400 = 2 : 1.

Example 8. A photon has momentum 6.63×10−276.63 \times 10^{-27} kg m/s. Its wavelength?

Solution: λ=hp=6.63×10−346.63×10−27=10−7\lambda = \dfrac{h}{p} = \dfrac{6.63 \times 10^{-34}}{6.63 \times 10^{-27}} = 10^{-7} m = 100 nm.

Solved Examples - Work Function & Threshold

Example 9. Threshold frequency of a metal with ϕ0\phi_0 = 3.3 eV?

Solution: ν0=ϕ0h=3.3×1.6×10−196.63×10−34≈8.0×1014\nu_0 = \dfrac{\phi_0}{h} = \dfrac{3.3 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 8.0 \times 10^{14} Hz.

Example 10. Threshold wavelength for ϕ0\phi_0 = 2.48 eV?

Solution: λ0=12402.48=500\lambda_0 = \dfrac{1240}{2.48} = 500 nm — green light is the limit; longer wavelengths eject nothing.

Example 11. A metal has threshold wavelength 620 nm. Its work function?

Solution: ϕ0=1240620=2.0\phi_0 = \dfrac{1240}{620} = 2.0 eV.

Example 12. Will 500 nm light eject electrons from a metal with ϕ0\phi_0 = 2.28 eV?

Solution: Photon energy =1240500=2.48= \dfrac{1240}{500} = 2.48 eV >2.28> 2.28 eV → yes, with Kmax=0.20K_{max} = 0.20 eV.

Example 13. Same metal, 600 nm light?

Solution: E=1240600=2.07E = \dfrac{1240}{600} = 2.07 eV <2.28< 2.28 eV → no emission, at any intensity.

Solved Examples - Einstein's Equation & Stopping Potential

Example 14. Light of 4.0 eV photons strikes a metal (ϕ0\phi_0 = 2.5 eV). Find KmaxK_{max} and V0V_0.

Solution: Kmax=4.0−2.5=1.5K_{max} = 4.0 - 2.5 = 1.5 eV; V0=1.5V_0 = 1.5 V.

Example 15. Cut-off voltage is 1.5 V. KmaxK_{max} in joules? (NCERT 11.3)

Solution: Kmax=eV0=1.5×1.6×10−19=2.4×10−19K_{max} = eV_0 = 1.5 \times 1.6 \times 10^{-19} = 2.4 \times 10^{-19} J.

Example 16. ϕ0\phi_0 = 2.14 eV, incident frequency 6×10146 \times 10^{14} Hz. Stopping potential? (NCERT 11.2)

Solution: hν=(6.63×10−34)(6×1014)1.6×10−19=2.49h\nu = \dfrac{(6.63 \times 10^{-34})(6 \times 10^{14})}{1.6 \times 10^{-19}} = 2.49 eV; V0=2.49−2.14=0.35V_0 = 2.49 - 2.14 = 0.35 V.

Example 17. Maximum speed in Example 16?

Solution: vmax=2×0.35×1.6×10−199.1×10−31≈3.5×105v_{max} = \sqrt{\dfrac{2 \times 0.35 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx 3.5 \times 10^5 m/s.

Example 18. Threshold 3.3×10143.3 \times 10^{14} Hz; incident 8.2×10148.2 \times 10^{14} Hz. Cut-off voltage? (NCERT 11.6)

Solution: V0=h(ν−ν0)e=6.63×10−34×4.9×10141.6×10−19≈2.0V_0 = \dfrac{h(\nu - \nu_0)}{e} = \dfrac{6.63 \times 10^{-34} \times 4.9 \times 10^{14}}{1.6 \times 10^{-19}} \approx 2.0 V.

Example 19. ϕ0\phi_0 = 4.2 eV. Emission for 330 nm light? (NCERT 11.7)

Solution: E=1240330=3.76E = \dfrac{1240}{330} = 3.76 eV < 4.2 eV → no emission.

Example 20. 488 nm light gives V0V_0 = 0.38 V. Work function? (NCERT 11.9)

Solution: E=1240488=2.54E = \dfrac{1240}{488} = 2.54 eV; ϕ0=2.54−0.38=2.16\phi_0 = 2.54 - 0.38 = 2.16 eV.

Example 21. Frequency 7.21×10147.21 \times 10^{14} Hz gives vmax=6.0×105v_{max} = 6.0 \times 10^5 m/s. Threshold frequency? (NCERT 11.8)

Solution: Kmax=12(9.1×10−31)(6×105)2=1.64×10−19K_{max} = \frac{1}{2}(9.1 \times 10^{-31})(6 \times 10^5)^2 = 1.64 \times 10^{-19} J; ν0=ν−Kmaxh=7.21×1014−2.47×1014=4.74×1014\nu_0 = \nu - \dfrac{K_{max}}{h} = 7.21 \times 10^{14} - 2.47 \times 10^{14} = 4.74 \times 10^{14} Hz.

Example 22. Light of 300 nm on sodium (ϕ0\phi_0 = 2.28 eV): maximum kinetic energy?

Solution: E=1240300=4.13E = \dfrac{1240}{300} = 4.13 eV; Kmax=4.13−2.28=1.85K_{max} = 4.13 - 2.28 = 1.85 eV.

Solved Examples - Graphs & Slopes

Example 23. Slope of the V0V_0-ν\nu graph is 4.12×10−154.12 \times 10^{-15} V s. Find h. (NCERT 11.5)

Solution: slope =h/e⇒h=1.6×10−19×4.12×10−15=6.59×10−34= h/e \Rightarrow h = 1.6 \times 10^{-19} \times 4.12 \times 10^{-15} = 6.59 \times 10^{-34} J s.

Example 24. The V0V_0-ν\nu line for a metal crosses the ν\nu-axis at 5×10145 \times 10^{14} Hz. Work function?

Solution: ϕ0=hν0=(6.63×10−34)(5×1014)1.6×10−19≈2.07\phi_0 = h\nu_0 = \dfrac{(6.63 \times 10^{-34})(5 \times 10^{14})}{1.6 \times 10^{-19}} \approx 2.07 eV.

Example 25. Two metals A and B have V0V_0-ν\nu lines with x-intercepts 4×10144 \times 10^{14} Hz and 6×10146 \times 10^{14} Hz. Which has the larger work function, and how do the slopes compare?

Solution: ϕ0=hν0\phi_0 = h\nu_0 → B (larger intercept) has the larger work function; the slopes are equal (both h/eh/e) — the lines are parallel.

Example 26. For a metal, V0V_0 = 1.0 V at ν=6×1014\nu = 6 \times 10^{14} Hz and V0V_0 = 2.0 V at ν=8.4×1014\nu = 8.4 \times 10^{14} Hz. Extract h.

Solution: h=eΔV0Δν=1.6×10−19×1.02.4×1014=6.67×10−34h = \dfrac{e\Delta V_0}{\Delta\nu} = \dfrac{1.6 \times 10^{-19} \times 1.0}{2.4 \times 10^{14}} = 6.67 \times 10^{-34} J s.

Solved Examples - Intensity, Saturation & Counting

Example 27. Saturation current is 6 μ\muA at intensity I. At intensity 3I (same frequency)?

Solution: saturation current ∝ intensity → 18 μ\muA; stopping potential unchanged.

Example 28. A source is moved from 1.0 m to 0.5 m from the emitter. Effect on saturation current and V0V_0?

Solution: intensity ∝ 1/r21/r^2 → ×4; saturation current ×4; V0V_0 unchanged (frequency untouched).

Example 29. Monochromatic light delivers 101510^{15} photons per second; quantum efficiency for emission is 0.1%. Photocurrent?

Solution: electrons/s =1015×10−3=1012= 10^{15} \times 10^{-3} = 10^{12}; I=1012×1.6×10−19=1.6×10−7I = 10^{12} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-7} A = 0.16 μ\muA. (Real surfaces eject far fewer electrons than photons received — NCERT notes most photons are absorbed without emission.)

Example 30. 30 kV electrons strike a target. Minimum X-ray wavelength? (NCERT 11.1)

Solution: λmin=1240 eV nm30000 eV=0.0413\lambda_{min} = \dfrac{1240 \text{ eV nm}}{30000 \text{ eV}} = 0.0413 nm; max frequency ν=eVh=7.24×1018\nu = \dfrac{eV}{h} = 7.24 \times 10^{18} Hz.

Solved Examples - de Broglie Wavelengths

Example 31. Electron at 5.4×1065.4 \times 10^6 m/s: de Broglie wavelength? (NCERT Example 11.3)

Solution: p=(9.11×10−31)(5.4×106)=4.92×10−24p = (9.11 \times 10^{-31})(5.4 \times 10^6) = 4.92 \times 10^{-24} kg m/s; λ=6.63×10−344.92×10−24=0.135\lambda = \dfrac{6.63 \times 10^{-34}}{4.92 \times 10^{-24}} = 0.135 nm.

Example 32. Electron accelerated through 150 V: wavelength?

Solution: λ=1.227150=1.22712.25≈0.10\lambda = \dfrac{1.227}{\sqrt{150}} = \dfrac{1.227}{12.25} \approx 0.10 nm.

Example 33. Bullet 0.040 kg at 1.0 km/s: wavelength? (NCERT 11.10a)

Solution: p=40p = 40 kg m/s; λ=6.63×10−3440=1.7×10−35\lambda = \dfrac{6.63 \times 10^{-34}}{40} = 1.7 \times 10^{-35} m — unmeasurable.

Example 34. Ball 0.060 kg at 1.0 m/s: wavelength? (NCERT 11.10b)

Solution: λ=6.63×10−340.060=1.1×10−32\lambda = \dfrac{6.63 \times 10^{-34}}{0.060} = 1.1 \times 10^{-32} m.

Example 35. Dust particle 10−910^{-9} kg at 2.2 m/s: wavelength? (NCERT 11.10c)

Solution: p=2.2×10−9p = 2.2 \times 10^{-9} kg m/s; λ=6.63×10−342.2×10−9=3.0×10−25\lambda = \dfrac{6.63 \times 10^{-34}}{2.2 \times 10^{-9}} = 3.0 \times 10^{-25} m.

Example 36. Electron and proton at the same speed: wavelength ratio λe:λp\lambda_e : \lambda_p?

Solution: λ∝1/m\lambda \propto 1/m at equal v → λe:λp=mp:me=1836:1\lambda_e : \lambda_p = m_p : m_e = 1836 : 1.

Example 37. Electron and proton with equal kinetic energy: ratio?

Solution: λ∝1/m\lambda \propto 1/\sqrt{m} at equal K → λe/λp=1836≈43\lambda_e/\lambda_p = \sqrt{1836} \approx 43.

Example 38. Electron and proton with equal momentum: ratio?

Solution: λ=h/p\lambda = h/p → equal momentum, equal wavelength: 1 : 1.

Example 39. What accelerating voltage gives an electron λ\lambda = 0.1227 nm?

Solution: V=(1.2270.1227)2=100V = \left(\dfrac{1.227}{0.1227}\right)^2 = 100 V.

Example 40. An alpha particle (mass 4u, charge 2e) and a proton (mass u, charge e) are accelerated through the same potential V. Ratio λa:λp\lambda_a : \lambda_p?

Solution: λ=h2mqV∝1mq\lambda = \dfrac{h}{\sqrt{2mqV}} \propto \dfrac{1}{\sqrt{mq}}; λaλp=mpe4mp⋅2e=18=122\dfrac{\lambda_a}{\lambda_p} = \sqrt{\dfrac{m_p e}{4m_p \cdot 2e}} = \dfrac{1}{\sqrt{8}} = \dfrac{1}{2\sqrt{2}} — the alpha's wavelength is 222\sqrt{2} times shorter.

Solved Examples - Mixed Mastery

Example 41. Light of 200 nm on a metal gives V0V_0 = 2.0 V. Find the threshold wavelength.

Solution: E=1240200=6.2E = \dfrac{1240}{200} = 6.2 eV; ϕ0=6.2−2.0=4.2\phi_0 = 6.2 - 2.0 = 4.2 eV; λ0=12404.2≈295\lambda_0 = \dfrac{1240}{4.2} \approx 295 nm.

Example 42. The stopping potential doubles when the wavelength drops from 400 nm to 300 nm. Find ϕ0\phi_0.

Solution: eV0=1240400−ϕ0=3.1−ϕ0eV_0 = \dfrac{1240}{400} - \phi_0 = 3.1 - \phi_0 and 2eV0=1240300−ϕ0=4.13−ϕ02eV_0 = \dfrac{1240}{300} - \phi_0 = 4.13 - \phi_0 (in eV). Subtracting: eV0=1.03eV_0 = 1.03 eV, so ϕ0=3.1−1.03=2.07\phi_0 = 3.1 - 1.03 = 2.07 eV.

Example 43. A photon of 5.0 eV ejects an electron (ϕ0\phi_0 = 3.0 eV). The ejected electron's own de Broglie wavelength (at KmaxK_{max})?

Solution: Kmax=2.0K_{max} = 2.0 eV → treat like an electron 'through 2.0 V': λ=1.2272.0≈0.87\lambda = \dfrac{1.227}{\sqrt{2.0}} \approx 0.87 nm.

Example 44. What wavelength of light would give photoelectrons of zero kinetic energy from caesium (ϕ0\phi_0 = 2.14 eV)?

Solution: zero KmaxK_{max} means threshold: λ0=12402.14≈579\lambda_0 = \dfrac{1240}{2.14} \approx 579 nm — yellow light just barely fails/succeeds at the edge.

Example 45. Keeping intensity fixed, the frequency is raised from 6×10146 \times 10^{14} to 9×10149 \times 10^{14} Hz on caesium (ν0=5.16×1014\nu_0 = 5.16 \times 10^{14} Hz). By what factor does KmaxK_{max} change?

Solution: K∝(ν−ν0)K \propto (\nu - \nu_0): ratio =9−5.166−5.16=3.840.84≈4.6= \dfrac{9 - 5.16}{6 - 5.16} = \dfrac{3.84}{0.84} \approx 4.6 — nearly a five-fold jump, because KmaxK_{max} measures the EXCESS above threshold, not the frequency itself.