The first ionization energy, e, of a potassium atom is 0.696 aj. what is the wavelength of light, in nm, that is just sufficient to ionize a potassium atom? values for constants can be found here.

The formula to be used for this problem is as follows:

E = hc/λ, where h is the Planck's constant, c is the speed of light and λ is the wavelength. Also 1 aJ = 10⁻¹⁸ J

0.696*10⁻¹⁸ = (6.62607004*10⁻³⁴ m²·kg/s) (3*10⁸ m/s) / λ

Solving for λ,

λ = 2.656*10⁻⁷ m or 0.022656 nm