The difference between threshold wavelengths for two metal surfaces A and B having work function φ_A=9 eV and φ_B=4 · 5 eV in nm is:
{ Given, hc =1242 eV nm}

Solution

Physics sir · 3 steps

Threshold wavelength is hc/φ: find it for each metal in nm, then subtract.

1. Threshold condition. At the threshold wavelength, one photon carries exactly the work function. So φ = hc/λ₀, which gives λ₀ = hc/φ.

   λ_A = 1242/9 nm = 138 nm, λ_B = 1242/4.5 nm = 276 nm

2. Take the difference. The metal with the larger work function has the shorter threshold wavelength. The difference is 276 - 138 = 138 nm.
3. So. The difference between the threshold wavelengths is 138 nm, option (b). It equals λ_A here only because φ_A is exactly twice φ_B.

The idea to keep. Threshold wavelength is inversely proportional to work function. With hc = 1242 eV nm and φ in eV, λ₀ = 1242/φ comes out directly in nm.

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