Dual Nature of Matter: Step-by-Step Worked Examples (4)

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Question

Light of wavelength 400 nm400 \, \text{nm} falls on a metal of work function 2.0 eV2.0 \, \text{eV}. Find the maximum kinetic energy of emitted photoelectrons in eV.

Solution — Step by Step

The energy of a photon with wavelength λ\lambda in nm is conveniently:

Ephoton=1240λ(nm) eVE_\text{photon} = \frac{1240}{\lambda(\text{nm})} \, \text{eV}

For λ=400 nm\lambda = 400 \, \text{nm}: Ephoton=1240/400=3.1 eVE_\text{photon} = 1240/400 = 3.1 \, \text{eV}.

Ephoton=ϕ+KEmax⁡E_\text{photon} = \phi + KE_{\max}

where ϕ\phi is the work function. Rearranging:

KEmax⁡=Ephoton−ϕ=3.1−2.0=1.1 eVKE_{\max} = E_\text{photon} - \phi = 3.1 - 2.0 = 1.1 \, \text{eV}.

Maximum kinetic energy of photoelectrons: KEmax⁡=1.1 eVKE_{\max} = 1.1 \, \text{eV}.

Why This Works

The “1240 eV·nm” trick converts wavelength to photon energy in one division — no need to multiply hh and cc in SI units, then divide by 1.6×10−191.6 \times 10^{-19} to convert to eV. That saves 60 seconds per problem.

Einstein’s equation is just energy conservation: the photon’s energy goes partly into freeing the electron from the metal (work function) and partly into the electron’s kinetic energy.

Memory hook: hc=1240 eV⋅nmhc = 1240 \, \text{eV·nm}. Whenever the question gives wavelength in nm and work function in eV, this shortcut is the fastest route. Students who haven’t memorised it lose 90 seconds in JEE Main per such question.

Alternative Method — SI Units

We could compute E=hc/λE = hc/\lambda in joules: E=(6.6×10−34)(3×108)/(400×10−9)=4.95×10−19 JE = (6.6 \times 10^{-34})(3 \times 10^8)/(400 \times 10^{-9}) = 4.95 \times 10^{-19} \, \text{J}. Convert to eV by dividing by 1.6×10−191.6 \times 10^{-19}: ≈3.1 eV\approx 3.1 \, \text{eV}. Same answer, but slower.

For JEE Main MCQs, always work in eV-nm. For derivations and Advanced subjective questions, SI units are sometimes safer.

Common Mistake

Students sometimes confuse work function with threshold wavelength and try to “subtract” wavelengths. Energy and wavelength are inversely related — we cannot subtract them directly.

Another trap: forgetting to convert ϕ\phi to the same units as EphotonE_\text{photon}. If the work function is given in joules and photon energy is computed in eV, we must convert one before subtracting. The “1240 trick” sidesteps this by keeping everything in eV.

JEE Main 2023 (Shift 2, January 31) used the same template with λ=248 nm\lambda = 248 \, \text{nm} and ϕ=3 eV\phi = 3 \, \text{eV}. The “1240/248 = 5 eV” computation is fast, then 5−3=2 eV5 - 3 = 2 \, \text{eV}. NEET asks 1-2 photoelectric problems every year — pure scoring topic.

The follow-up usually asks for stopping potential (V0=KEmax⁡/e=1.1 VV_0 = KE_{\max}/e = 1.1 \, \text{V}) or maximum velocity. Both are one-liner extensions.

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