Dual Nature of Matter: Tricky Questions Solved (3)

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Question

Light of wavelength 400 nm400 \text{ nm} falls on a metal of work function ϕ=2.0 eV\phi = 2.0 \text{ eV}. Find: (a) the maximum kinetic energy of photoelectrons, (b) the de Broglie wavelength of the most energetic photoelectron, (c) the stopping potential.

Use h=6.63×10−34h = 6.63 \times 10^{-34} J·s, c=3×108c = 3 \times 10^8 m/s, me=9.11×10−31m_e = 9.11 \times 10^{-31} kg, 1 eV=1.6×10−191 \text{ eV} = 1.6 \times 10^{-19} J.

Solution — Step by Step

A handy shortcut: E(eV)=1240λ(nm)E(\text{eV}) = \tfrac{1240}{\lambda(\text{nm})}. So:

E=1240400=3.1 eVE = \tfrac{1240}{400} = 3.1 \text{ eV} KEmax⁡=E−ϕ=3.1−2.0=1.1 eVKE_{\max} = E - \phi = 3.1 - 2.0 = 1.1 \text{ eV}

In SI: KEmax⁡=1.1×1.6×10−19=1.76×10−19KE_{\max} = 1.1 \times 1.6 \times 10^{-19} = 1.76 \times 10^{-19} J.

For non-relativistic electrons, p=2meKEp = \sqrt{2m_e KE}:

p=2×9.11×10−31×1.76×10−19p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.76 \times 10^{-19}} p≈3.21×10−49≈5.66×10−25 kg⋅m/sp \approx \sqrt{3.21 \times 10^{-49}} \approx 5.66 \times 10^{-25} \text{ kg·m/s} λe=hp=6.63×10−345.66×10−25≈1.17×10−9 m=1.17 nm\lambda_e = \tfrac{h}{p} = \tfrac{6.63 \times 10^{-34}}{5.66 \times 10^{-25}} \approx 1.17 \times 10^{-9} \text{ m} = 1.17 \text{ nm}

By definition eV0=KEmax⁡eV_0 = KE_{\max}, so:

V0=KEmax⁡e=1.1 VV_0 = \tfrac{KE_{\max}}{e} = 1.1 \text{ V}

Final answers: KEmax⁡=1.1 eVKE_{\max} = \mathbf{1.1 \text{ eV}}, λe≈1.17 nm\lambda_e \approx \mathbf{1.17 \text{ nm}}, V0=1.1 VV_0 = \mathbf{1.1 \text{ V}}.

Why This Works

The Einstein photoelectric equation hν=ϕ+KEmax⁡h\nu = \phi + KE_{\max} is just energy conservation — incoming photon energy goes into ejecting the electron (work function) plus the leftover kinetic energy.

The de Broglie relation λ=h/p\lambda = h/p then connects this kinetic energy to a wavelength, showcasing wave-particle duality: photons hit the metal as light and the ejected electrons travel as matter waves.

Alternative Method

The shortcut λe(nm)≈1.226KE(eV)\lambda_e(\text{nm}) \approx \tfrac{1.226}{\sqrt{KE(\text{eV})}} gives the de Broglie wavelength directly. Plugging in KE=1.1KE = 1.1 eV: λe≈1.2261.049≈1.17\lambda_e \approx \tfrac{1.226}{1.049} \approx 1.17 nm. Saves a minute in JEE.

Common Mistake

Using the photon’s wavelength as the electron’s de Broglie wavelength. They’re entirely different quantities. The photon at 400400 nm gave the electron only 1.11.1 eV of KE — the electron’s matter-wave length is 1.171.17 nm, much shorter than the photon wavelength.

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