Dual Nature of Matter: Application Problems (5)

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Question

Light of wavelength λ=400\lambda = 400 nm falls on a metal of work function ϕ0=2.0\phi_0 = 2.0 eV. Find the maximum kinetic energy of the photoelectrons and the stopping potential. (h=6.63×10−34h = 6.63 \times 10^{-34} J·s, c=3×108c = 3 \times 10^8 m/s, 11 eV =1.6×10−19= 1.6 \times 10^{-19} J.)

Solution — Step by Step

E=hcλ=(6.63×10−34)(3×108)400×10−9E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{400 \times 10^{-9}}

E=1.989×10−254×10−7=4.97×10−19 JE = \frac{1.989 \times 10^{-25}}{4 \times 10^{-7}} = 4.97 \times 10^{-19} \text{ J}

Convert to eV: E=4.97×10−19/1.6×10−19≈3.1E = 4.97 \times 10^{-19} / 1.6 \times 10^{-19} \approx 3.1 eV.

Quick shortcut: EE (in eV) =1240/λ= 1240/\lambda (in nm) =1240/400=3.1= 1240/400 = 3.1 eV. ✓

Kmax=E−ϕ0=3.1−2.0=1.1 eVK_{max} = E - \phi_0 = 3.1 - 2.0 = 1.1 \text{ eV}

eV0=Kmax  ⟹  V0=1.1 VeV_0 = K_{max} \implies V_0 = 1.1 \text{ V}

(When you express KmaxK_{max} in eV, the stopping potential is numerically equal in volts. Slick.)

Photon energy (3.13.1 eV) > work function (2.02.0 eV), so emission happens. Excess energy (1.11.1 eV) appears as kinetic energy of the fastest electrons.

Final answer: Kmax=1.1K_{max} = 1.1 eV, V0=1.1V_0 = 1.1 V.

Why This Works

Einstein’s 1905 explanation: light is quantised into photons of energy hνh\nu. A photon either gives all its energy to one electron or none — no partial absorption. So Ephoton=ϕ0+KmaxE_{photon} = \phi_0 + K_{max}.

The work function ϕ0\phi_0 is the minimum energy needed to liberate an electron from the metal surface. If Ephoton<ϕ0E_{photon} < \phi_0, no emission at all, regardless of intensity. This kills the classical wave picture, where intensity should determine emission.

Photon energy: E=hν=hc/λE = h\nu = hc/\lambda

Quick conversion: EE (eV) =1240/λ= 1240/\lambda (nm)

Einstein’s equation: Kmax=E−ϕ0K_{max} = E - \phi_0

Threshold frequency: ν0=ϕ0/h\nu_0 = \phi_0/h (below this, no emission)

Stopping potential: eV0=KmaxeV_0 = K_{max}, so V0=Kmax/eV_0 = K_{max}/e (or numerically equal if KK is in eV)

Alternative Method

Use the threshold wavelength directly. Threshold wavelength λ0=hc/ϕ0=1240/2.0=620\lambda_0 = hc/\phi_0 = 1240/2.0 = 620 nm.

Kmax=hc(1λ−1λ0)=1240(1400−1620)=1240⋅220248000≈1.1 eVK_{max} = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right) = 1240\left(\frac{1}{400} - \frac{1}{620}\right) = 1240 \cdot \frac{220}{248000} \approx 1.1 \text{ eV}

Same answer.

The graph of KmaxK_{max} vs ν\nu is a straight line with slope hh and x-intercept ν0\nu_0. JEE Main loves this graph — the slope is universal (Planck’s constant), but the x-intercept depends on the metal.

Common Mistake

Students confuse the stopping potential with the applied accelerating potential. Stopping potential is the negative voltage that just stops the fastest photoelectron — it equals Kmax/eK_{max}/e. Don’t try to find it from “the voltage in the circuit.”

Another trap: assuming photoelectric current depends on wavelength alone. Current depends on intensity (number of photons per second), while KmaxK_{max} depends on frequency (energy per photon). Two independent variables.

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