Practice question

Given: The threshold wavelength for a metal is 500 nm . What is the work function of the metal in eV? (Take h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J ) These values define the system as per NCERT data. Formula: Threshold frequency v_0 = c/lambda_0 = frac3 × 10⁸⁵⁰⁰ × 10⁻⁹= 6 × 10¹⁴ Hz. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: phi_0 = h v_0 = 6.63 × 10⁻³⁴ × 6 × 10¹⁴= 3.978 × 10⁻¹⁹ J . phi_0 = frac3.978 × 10⁻¹⁹¹.6 × 10⁻¹⁹ approx 2.49 eV . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.