When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy T A eV end de-Broglie wavelength λ _A. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is T B = (T A – 1.5) eV. If the de-Broglie wavelength of these photoelectrons λ _B = 2λ _A, then the work function of metal B is :

Solution

Physics sir · 4 steps

De Broglie wavelength goes as one over root of kinetic energy, so doubling λ cuts T to a quarter.

1. Wavelength and energy. For an electron with kinetic energy T, the momentum is p = √2mT. So λ = h/√2mT, which means λ ∝ 1/√T.

   λ_B/λ_A = √(T_A/T_B) = 2 ⇒ T_B = T_A/4

2. Find the energies. Set this equal to the given relation: T_A - 1.5 = T_A/4. Then 3T_A/4 = 1.5, so T_A = 2 eV and T_B = 0.5 eV.
3. Einstein's equation. For metal B, φ_B = E - T_B = 4.5 - 0.5 = 4 eV.
4. So. The work function of metal B is 4 eV, option (b).

The idea to keep. For a particle with kinetic energy T, use λ = h/√2mT. A ratio of wavelengths gives a ratio of energies through a square root, so square it before using it.

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