When electromagnetic radiation of wavelength 300 ~nm falls on the surface of a metal, electrons are emitted with the kinetic energy of 1.68 10^5 J mol ⁻¹ . What is the minimum energy needed to remove an electron from the metal? aligned & (h=6.626 10⁻³⁴ Js , c =3 10^8 ~ms ⁻¹, . & N _ A =6.022 10²³ ~mol ⁻¹ aligned

Step-by-step solution

Since, the striking photon has energy equal to h v v and the minimum energy required to eject the electron is h v₀ (work function W ₀ ), the difference in energy (h v-h v₀ ) is transferred as the kinetic energy of the photoelectron. h v=h v₀+ KE h v₀= minimum energy required to remove electron) Energy of a 300 ~nm photon is given by aligned & E = h c & = 6.626 10⁻³⁴ Js 3 10^8 ~ms ⁻¹ 300 10⁻⁹ ~m & =6.626 10⁻¹⁹ ~J aligned Energy of one mole of photons aligned & =6.626 10⁻¹⁹ ~J 6.022 10²³ ~mol ⁻¹ & =3.99 10^5 ~J ~mol