Practice question

Given: A metal has a threshold frequency of 4.5 × 10¹⁴ Hz . What is the maximum kinetic energy of electrons emitted by light of frequency 6.0 × 10¹⁴ Hz ? (Take h = 6.63 × 10⁻³⁴ J s ) These values define the system as per NCERT data. Formula: phi_0 = h v_0 = 6.63 × 10⁻³⁴ × 4.5 × 10¹⁴= 2.9835 × 10⁻¹⁹ J. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E = h v = 6.63 × 10⁻³⁴ × 6.0 × 10¹⁴= 3.978 × 10⁻¹⁹ J . K_{max = E - phi_0 = 3.978 × 10⁻¹⁹- 2.9835 × 10⁻¹⁹= 9.945 × 10⁻²⁰ J . 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.