Question

When the radiation of 58.4 nm from a helium discharge lamp is directed on a sample of krypton, electrons are ejected with a velocity of 1.59 x 10^6m s^-1. THe same radiation ejects electrons from rubidium atoms with a velocity of 2.45 x 10^6m s^-1. What are the ionization energies (in eV) of the two elements?

Answer 1

krypton

Energy of radiation with wavelength 58.4 nm is E = h c / λ

Where

E = (6.625 x 10 ^-34 x 3 x10 ⁸ ) / (  58.4 x 10 ^-9 )

       = 34.032 x 10 ^-19 J

velocity of v = 1.59 x 10 ⁶ m s ^-1

So Kinetic energy of ejected electron is ,

K.E = (1/2) mv ²  

        = 0.5 x 9.11 x 10 ^-31 Kg x (1.59 x 10 ⁶ m s ^-1) ²

         = 1.15 x 10 ^-18 J

        = 11.5 x 10 ^-19 J

We know that ,

the energy of the photon(radiation) = K.E of e- + work function(or ionization energy)

                      34.032 x 10 ^-19 J    = 11.5 x 10 ^-19 J +  ionization energy

So ionization energy of Kr is = 22.532 x 10 ^-19 J

For Rb atoms ,

Kinetic energy of ejected electron is ,

                  K.E = (1/2) mv ²

                         = 0.5 x 9.11 x 10 ^-31 Kg x (2.45 x 10 ⁶ m s ^-1) ²

                         = 2.734 x 10 ^-18 J

                         = 27.34 x 10 ^-19 J

We know that ,

the energy of the photon(radiation) = K.E of e- + work function(or ionization energy)

                      34.032 x 10 ^-19 J    = 27.34 x 10 ^-19 J +  ionization energy

So  ionization energy of Rb is = 6.692 x 10 ^-19 J