The minimum energy required to eject an electron from the surface of a specific metal is 4.15 10^5 J mol ⁻¹ . If the metal surface is exposed to electromagnetic radiation of wavelength 200 nm , what will be the maximum kinetic energy of the emitted photoelectrons per mole? (h = 6.626 10⁻³⁴ Js , c = 3 10^8 ms ⁻¹, N_A = 6.022 10²³ mol ⁻¹ )

Step-by-step solution

The energy of one mole of incident photons is given by E = hcN_A Substituting the given values: E = 6.626 10⁻³⁴ 3 10^8 6.022 10²³ 200 10⁻⁹ E = 19.878 10⁻²⁶ 6.022 10²³ 2 10⁻⁷ E = 9.939 10⁻¹⁹ 6.022 10²³ E = 59.85 10^4 J mol ⁻¹ = 5.985 10^5 J mol ⁻¹ According to Einstein's photoelectric equation, the maximum kinetic energy is: K_ max = E - W K_ max = 5.985 10^5 - 4.15 10^5 K_ max = 1.835 10^5 J mol ⁻¹ Rounding off to three significant figures, we get 1.84 10^5 J mol ⁻¹ . Answer: 1.84 10^5 J mol ⁻¹