If the two metals A and B are exposed to radiation of wavelength 350 nm. The work functions of metals A and B are 4.8 eV and 2.2 eV. Then choose the correct option.

Solution

Physics sir · 2 steps

Turn 350 nm into electron-volts first, then check it against each metal's work function.

1. Condition for emission. A metal emits photoelectrons only if one photon carries at least the work function: Eₚₕₒₜₒₙ ≥ φ. Brightness does not matter. One photon frees one electron.

   E = hc/λ = (1240 eV nm)/(350 nm) ≈ 3.54 eV

   Metal	Work function	Photon energy	Emits?
   A	4.8 eV	3.54 eV	No, since 3.54 < 4.8
   B	2.2 eV	3.54 eV	Yes, since 3.54 > 2.2

2. So. Metal B emits photoelectrons, with maximum kinetic energy about 3.54 - 2.2 = 1.34 eV. Metal A emits none, so the answer is (c).

The idea to keep. Remember hc ≈ 1240 eV nm, so the photon energy in eV is 1240 divided by the wavelength in nm. Electrons come out only when this beats the work function.

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