Question

Using Balmer series, calculate ionisation energy of hydrogen atom in kJ mol1

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Answer 1

For this we use the Rydberg formula (generalized Balmer series formula).

    1/λ = 4/B (1/2² - 1/n²) =  R (1/2² - 1/n²)
       R = 1.097 *10⁷ m⁻¹
    Ionization energy corresponds to λ = 1/R = 91.1 nm
    Energy = h c/λ = 21.92 * 10⁻¹⁹ J /atom
                  = 1312 kJ/mol   by multiplying with Avogadro number
         ( h = Planck's constant = 6.62 * 10^-34 units)

             c = speed of light
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Balmer series spectral lines formula :(for transitions from n = 2)
 
   λ = B * n² / (n² - 2²),              B = 364.5 nm
       for wavelength of radiation emission for a   n -> 2 transition.

For ionization, an electron changes from n=2, to n = ∞.

Hence,  ionization energy = E = h c /λ 
    = 6.62 * 10⁻³⁴ * 3 * 10⁸ / 364.5 * 10⁻⁹  J
    = 5.45 * 10⁻¹⁹ J / atom or electron
    = 328.4 kJ / mole    (by multiplying with Avogadro Number).

This value seems to be 1/4 of the 1312 kJ/mole value observed actually (for transition from n = 1 to infinity).

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Ionization energy is found also found by finding the frequency series limit of the Hydrogen spectrum. It requires extrapolation of the frequency & energy.

   It is found that the frequency is = 3.28 * 10¹⁵ Hz at the limit that corresponds to the ionization. Because above this frequency there are no emissions or absorptions.

   Ionization energy = 3.28 * 10¹⁵ * 6.62 * 10⁻³⁴ * 6.022 * 10²³
                                 = 1308 kJ/mole