Practice question

Given: The work function of a metal is 3.0 eV . Light of frequency 8.0 × 10¹⁴Hz is incident on it. What is the stopping potential? (Take h = 6.63 × 10⁻³⁴J s, e = 1.6 × 10⁻¹⁹C ) These values define the system as per NCERT data. Formula: E = h v = 6.63 × 10⁻³⁴ × 8.0 × 10¹⁴= 5.304 × 10⁻¹⁹J. This is standard NCERT relation. Substitution & Calculation: E = frac5.304 × 10⁻¹⁹¹.6 × 10⁻¹⁹approx 3.315 eV . K_{max = E - phi_0 = 3.315 - 3.0 = 0.315 eV . V_0 = fracK_{maxe = 0.315 V . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.