In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2 V , what is the wavelength of the em-wave? (Given hc=1242 eVnm where h is the Planck's constant and c is the speed of light in vaccum.)

Solution

Physics sir · 4 steps

The photon pays the work function first, and the stopping potential tells you what is left over.

1. Energy balance. Each photon gives all its energy to one electron. Part of it frees the electron (the work function φ), and the rest is the maximum kinetic energy, which equals eV₀.

   hc/λ = φ + eV₀

2. Photon energy. A stopping potential of 2 V means eV₀ = 2 eV. So the photon energy is 2.14 + 2 = 4.14 eV.
3. Find wavelength. λ = hc/E = (1242 eV nm)/(4.14 eV) = 300 nm.
4. So. The incident wave has wavelength 300 nm, which is option (c).

The idea to keep. Photon energy = work function + eV₀. In eV, a stopping potential of V₀ volts is V₀ eV of kinetic energy, and hc = 1242 eV nm turns energy into wavelength in one step.

Answer from the NTA final answer key. Solution by PrepWiser.