JEE Main 2023 · 12 April · morning · Numerical answer · +4 / −0
Values of work function (W₀) for a few metals are given below
The number of metals which will show photoelectric effect when light of wavelength 400 nm falls on it is ___________
Given: h=6.6 × 10⁻³⁴ J s
c=3 × 10⁸ ms⁻¹
e=1.6 × 10⁻¹⁹ C
| Metal | Li | Na | K | Mg | Cu | Ag |
|---|---|---|---|---|---|---|
| W₀/eV | 2.42 | 2.3 | 2.25 | 3.7 | 4.8 | 4.3 |
Answer and solution
Answer 3
Solution
Chemistry ma’am · 3 steps
Find the photon energy in eV once, then any metal whose work function is lower than it will emit electrons.
- Condition for emission. A metal shows the photoelectric effect only when one photon carries at least the work function: E ≥ W₀. Intensity does not matter, so compare E with each W₀.
E = hc/λ = (6.6 × 10⁻³⁴ × 3 × 10⁸)/(400 × 10⁻⁹) = 4.95 × 10⁻¹⁹ J
- Convert to eV. Divide by e: E = (4.95 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) ≈ 3.09 eV. This puts it in the same units as the table.
Metal W₀ / eV Is W₀ < 3.09 eV? Photoelectric effect Li 2.42 yes shown Na 2.3 yes shown K 2.25 yes shown Mg 3.7 no not shown Cu 4.8 no not shown Ag 4.3 no not shown - So. Only Li, Na and K have work functions below 3.09 eV, so the number of metals is 3.
The idea to keep. Photoelectric questions with a list of metals come down to one number: turn hc/λ into eV, then count the work functions below it. The quick check is E(eV) ≈ 1240/λ(nm).
Answer from the NTA final answer key. Solution by PrepWiser.