Electromagnetic radiation of wavelength 663 nm is just sufficient to ionise the atom of metal A. The ionization enegy of metal A in kJ mol⁻¹ is __________. (Rounded off to the nearest integer)
[h = 6.63 × 10⁻³⁴ Js, c = 3.00 × 10⁸ ms⁻¹, N A = 6.02 × 10²³ mol⁻¹]

Solution

Chemistry ma’am · 3 steps

Find the energy of one photon, then multiply by Avogadro's number to get a mole's worth.

1. Just sufficient. "Just sufficient to ionise" means one photon carries exactly the energy needed to remove one electron from one atom. So the photon energy equals the ionisation energy per atom.

   E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(663 × 10⁻⁹) J = 3.00 × 10⁻¹⁹ J

2. Per mole. One mole of atoms needs N_A such photons: 3.00 × 10⁻¹⁹ × 6.02 × 10²³ = 1.806 × 10⁵ J mol⁻¹.
3. So. The ionisation energy is 180.6 kJ mol⁻¹, which rounds to 181.

The common slip: 180600. Stopping at J mol⁻¹ gives 180600. Divide by 1000 to report the answer in kJ mol⁻¹.

The idea to keep. Photon energy hc/λ is for one atom. Whenever the answer is asked per mole, multiply by N_A, and convert nm to m before you divide.

Answer from the NTA final answer key. Solution by PrepWiser.