Question

An X-ray photon of wavelength 0.958 nm strikes a surface. The emitted electron has a kinetic energy of 978 eV .

What is the binding energy of the electron in kJ mol−1?
[KE = 12mv2; 1 electron volt (eV) = 1.602×10−19J]

Answer 1

E   = hc/

    = 0.958nm   = 0.958*10^-9m =

h = 6.625*10^-34 Joule-sec

C = 3*10^8 m/sec

Ephoton = 6.625*10^-34*3*10^8/0.958*10^-9    = 2.075*10^-16 J

E electron   = 978ev

                     = 978*1.602*10^-19J

                     = 1.56*10^-16J

E binding energy   = E photon - E electron

                               = 2.075*10^-16 - 1.56*10^-16

                               = (2.075-1.56)*10^-16

                                = 5.15*10^-17 J     = 5.15*10^-20Kj

energy converted to KJ to KJ/mole

               E binding energy   = 5.15*10^-20KJ*6.023*10^23/moles        = 3.1*10^4KJ/mole