Question

The first ionization energy of thallium is ei1 = 589.3 kj/mol. What is the longest wavelength of light that could ionize an atom of thallium?

Answer 1

Answer : The longest wavelength of light that could ionize an atom of thallium is  2.0314 × $10^{-7}$ m.

Solution : First we will calculate  the ionization energy for 1 atom of thallium.

As we know that, one mole of any element contains 6.022 × $10^{23}$ atoms as given by the Avogadro's number.

The energy needed to ionize 1 atom of of thallium = $\frac{589.3 \text{ KJ/mol}}{6.022\times 10^{23}\text{ atoms}}\times 1 \text{ mole of thallium}$

                                                                                     = 97.8578 × $10^{-23}$

                                                                                     = 9.7857 × $10^{-22}$ KJ/atom

As we know that the energy of a photon is directly proportional to the frequency which is given by the Planck equation.

       E =  h  ×  ν  

       ν =  $\frac{E}{h}$          ........... (1)

here,   E → the energy of the photon

           ν → frequency

           h → Planck's constant is equal to 6.626 × $10^{-34}$ JS

Put the values of E and h in equation (1) , we get

    $\nu=\frac{E}{h}$ = $\frac{9.7857 \times10 ^{-22}\times 10^{3} \text{ J}}{6.626\times 10^{-34}\text{ JS}}$

                = 1.4768 × $10^{15}\ S^{-}$

Now the frequency of a photon is inversely proportional to the wavelength .i.e

              $\nu = \frac{C}{\lambda}$      ............(2)

 here,   λ  →   wavelength of photon

            C  →   speed of light is equal to 3 × $10^{8} \text{ m} S^{-}$

 Now put the values of of C and λ in equation (2) , we get

          $\lambda = \frac{3\times 10^{8} \text{ m} S^{-}}{1.4768 \times  10^{15}S^{-}}$

                                 = 2.0314 × $10^{-7}$ m