Respuesta :
Answer : The longest wavelength of light that could ionize an atom of thallium is 2.0314 × [tex]10^{-7}[/tex] 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 × [tex]10^{23}[/tex] atoms as given by the Avogadro's number.
The energy needed to ionize 1 atom of of thallium = [tex]\frac{589.3 \text{ KJ/mol}}{6.022\times 10^{23}\text{ atoms}}\times 1 \text{ mole of thallium}[/tex]
= 97.8578 × [tex]10^{-23}[/tex]
= 9.7857 × [tex]10^{-22}[/tex] 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 = [tex]\frac{E}{h}[/tex] ........... (1)
here, E → the energy of the photon
ν → frequency
h → Planck's constant is equal to 6.626 × [tex]10^{-34}[/tex] JS
Put the values of E and h in equation (1) , we get
[tex]\nu=\frac{E}{h}[/tex] = [tex]\frac{9.7857 \times10 ^{-22}\times 10^{3} \text{ J}}{6.626\times 10^{-34}\text{ JS}}[/tex]
= 1.4768 × [tex]10^{15}\ S^{-}[/tex]
Now the frequency of a photon is inversely proportional to the wavelength .i.e
[tex]\nu = \frac{C}{\lambda}[/tex] ............(2)
here, λ → wavelength of photon
C → speed of light is equal to 3 × [tex]10^{8} \text{ m} S^{-}[/tex]
Now put the values of of C and λ in equation (2) , we get
[tex]\lambda = \frac{3\times 10^{8} \text{ m} S^{-}}{1.4768 \times 10^{15}S^{-}}[/tex]
= 2.0314 × [tex]10^{-7}[/tex] m
