Modern Physics: Application Problems (1)

easy 2 min read

Question

The work function of a metal is ϕ=2.5\phi = 2.5 eV. Light of wavelength λ=400\lambda = 400 nm falls on the surface. Find the maximum kinetic energy of photoelectrons. Take hc=1240hc = 1240 eV nm.

Solution — Step by Step

Using E=hc/λE = hc/\lambda with hc=1240hc = 1240 eV nm:

E=1240400=3.1 eVE = \frac{1240}{400} = 3.1 \text{ eV}

This is the trick that saves time — keep hchc in eV nm and λ\lambda in nm; the answer comes out in eV directly.

KEmax=EphotonϕKE_{\max} = E_{\text{photon}} - \phi

KEmax=3.12.5=0.6 eVKE_{\max} = 3.1 - 2.5 = 0.6 \text{ eV}

Final answer: KEmax=0.6KE_{\max} = 0.6 eV.

If the question asked for the stopping potential, that’s just V0=KEmax/e=0.6V_0 = KE_{\max}/e = 0.6 V.

Why This Works

Einstein’s photoelectric equation: each photon delivers energy hνh\nu to a single electron. Part of it (ϕ\phi, the work function) goes into freeing the electron from the metal; the rest becomes kinetic energy. Photons below threshold (E<ϕE < \phi) cause no emission, regardless of intensity.

The threshold wavelength is λ0=hc/ϕ\lambda_0 = hc/\phi. For this metal, λ0=1240/2.5=496\lambda_0 = 1240/2.5 = 496 nm. Since 400<496400 < 496, emission happens.

Alternative Method

Compute everything in joules: E=hc/λ=(6.63×1034)(3×108)/(400×109)4.97×1019E = hc/\lambda = (6.63 \times 10^{-34})(3 \times 10^8)/(400 \times 10^{-9}) \approx 4.97 \times 10^{-19} J. Convert work function: ϕ=2.5×1.6×1019=4×1019\phi = 2.5 \times 1.6 \times 10^{-19} = 4 \times 10^{-19} J. Then KE=9.7×1020KE = 9.7 \times 10^{-20} J =0.6= 0.6 eV. Slower but mechanically equivalent.

Memorise hc=1240hc = 1240 eV nm. With this single number, you can compute photon energies in your head for any visible-light wavelength. JEE timing rewards this kind of shortcut.

Common Mistake

Students sometimes write KE=hν+ϕKE = h\nu + \phi instead of KE=hνϕKE = h\nu - \phi. The work function is the energy COST to escape, not a bonus — so it subtracts. Also: KEmaxKE_{\max} is the maximum, since some electrons lose energy to the metal lattice on the way out.

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