Determine the ionization energy in kilojoules per mole of a

Determine the ionization energy (in kilojoules per mole) of a metallic element if light of wavelength 67.2 nm ejects electrons with a speed of 2165 km/sec.

Solution

hc/lamda=1/2mv^2+ ionization energy

ionization energy=hc/lamda+1/2mv^2

=[6.624*10-34*3*10^8]/67.2*10-9-0.5*9.1*10^-31*[2165*1000]^2

=2.957*10^-18-2.132*10-18

=8.243*10^-19J