What is the ionisation energy of hydrogen atom in Balmer series?
Answers
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
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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