The first ionization energy of a potassium atom is 0696 aJ a

The first ionization energy of a potassium atom is 0.696 aJ (attojoules). What is the frequency and wavelength, in nanometers, of photons capable of just ionizing potassium atoms? Values for constants can be found here. Assuming an ionization efficiency of 35.0%, how many such photons are needed to ionize 1.00 times 10^16 atoms?

Solution

E = h v

v = frequency = 0.696 * 10^-18 / 6.626 * 10^-34

= 1.05 * 10¹⁵ s^-1

E = h c / lambda

lambda = (6.626 * 10^-34 * 3 * 108) / (0.696 * 10^-18)

= 285 nm

number of photons = 1 * 10¹⁶ / 0.35

= 2.86 * 10¹⁶