Practice question

Given: The maximum speed of photoelectrons emitted from a surface is 5.0 × 10⁵m/s . What is the stopping potential? (Take m_e = 9.11 × 10⁻³¹kg, e = 1.6 × 10⁻¹⁹C ) Formula: K_{max = 1/2 m v_{max² = 1/2 × 9.11 × 10⁻³¹ × (5.0 × 10⁵)² = 1.13875 × 10⁻¹⁹J. Substitution & Calculation: V_0 = fracK_{maxe = frac1.13875 × 10⁻¹⁹¹.6 × 10⁻¹⁹approx 0.71 V . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.