The minimum energy required to remove an electron from the surface of a metal is 2.50 10^5 J mol ⁻¹ . When electromagnetic radiation falls on this metal, photoelectrons are emitted with a maximum kinetic energy of 1.49 10^5 J mol ⁻¹ . What is the wavelength of the incident radiation? (h = 6.626 10⁻³⁴ Js , c = 3 10^8 ms ⁻¹, N_A = 6.022 10²³ mol ⁻¹)

Step-by-step solution

According to Einstein's photoelectric equation, the total energy of the incident radiation is the sum of the work function and the maximum kinetic energy of the emitted electrons. Total energy per mole of incident photons: E_ mol = W₀ + KE E_ mol = (2.50 10^5 J mol ⁻¹) + (1.49 10^5 J mol ⁻¹) = 3.99 10^5 J mol ⁻¹ Energy of a single incident photon: E = E_ mol N_A E = 3.99 10^5 J mol ⁻¹ 6.022 10²³ mol ⁻¹ = 6.626 10⁻¹⁹ J Using Planck's quantum theory, E = hc = hc E = 6.626 10⁻³⁴ Js 3 10^8 ms ⁻¹ 6.626 10⁻¹⁹ J = 3 10⁻⁷