Practice question

Given: Light of frequency 5.0 × 10¹⁴ Hz is incident on a metal surface with work function 1.8 eV . What is the maximum speed of emitted electrons? (Take h = 6.63 × 10⁻³⁴ J s, m_e = 9.11 × 10⁻³¹ kg ) These values define the system as per NCERT data. Formula: E = h v = 6.63 × 10⁻³⁴ × 5.0 × 10¹⁴= 3.315 × 10⁻¹⁹ J. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E = frac3.315 × 10⁻¹⁹¹.6 × 10⁻¹⁹ approx 2.07 eV . K_{max = E - phi_0 = 2.07 - 1.8 = 0.27 eV = 0.27 × 1.6 × 10⁻¹⁹= 4.32 × 10⁻²⁰ J . K_{max = 1/2 m v_{max² Rightarrow v_{max = sqrtfrac2 K_{maxm = sqrtfrac2 × 4.32 × 10⁻²⁰⁹.11 × 10⁻³¹ approx 3.08 × 10⁵ m/s . 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.