Light of frequency 7.5 × 10¹⁴Hz produces a stopping potential of 0.9 V . What is the work function in eV? (Take h = 6.63
Given: Light of frequency 7.5 × 10¹⁴Hz produces a stopping potential of 0.9 V . What is the work function in eV? (Take h = 6.63 × 10⁻³⁴J s ) Formula: E = h v = 6.63 × 10⁻³⁴ × 7.5 × 10¹⁴= 4.9725 × 10⁻¹⁹J. Substitution & Calculation: E = frac4.9725 × 10⁻¹⁹¹.6 × 10⁻¹⁹approx 3.108 eV . K_{max = e V_0 = 0.9 eV . phi_0 = E - K_{max = 3.108 - 0.9 approx 2.208 eV . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.
Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.
