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It’s a well known fact that protium,deuterium and tritium have the same number of electrons and protons but different number of neutrons. It is expected that their ionisation enthalpies should be same as in all the cases only one proton attracts the only electron they have but their enthalpies are different.

What causes this?

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    $\begingroup$ Because the presence of the neutrons changes the nuclear spin state and required wave function symmetry. $\endgroup$ – Jon Custer Feb 1 '17 at 14:27
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    $\begingroup$ there are probably a handful of corrections needed to calculate the exact difference. But the biggest factor is just the mass of the nucleus. Not only is the electron attracted to the nucleus, the nucleus is attracted to the electron. So not only does the electron "move", so does the nucleus. A heavier nucleus moves less. This in turn alters the wave function of the electron. $\endgroup$ – MaxW Feb 1 '17 at 15:29
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There are a handful of additional corrections needed to calculate the exact differences. But the biggest factor is just the mass of the nucleus. Not only is the electron attracted to the nucleus, the nucleus is attracted to the electron. So not only does the electron "move", so does the nucleus. A heavier nucleus moves less. This in turn alters the wave function of the electron.

The difference between protium, deuterium and tritium is reasonably explained by the Bohr model of the atom. The gist is that the electron does not travel around a nucleus of infinite mass, but rather that the electron and nucleus travel around their common center of mass.

For one electron traveling around the nucleus the energy levels of the Bohr system are:

$\text{E}_n = \dfrac{1}{1+ \dfrac{\text{M}_e}{\text{M}}} \cdot \dfrac{\text{Z}^2}{\text{n}^2}\cdot \text{E}_R$

$\text{E}_n$ Energy of the $n$th orbital of the one electron system
$\text{M}_e$ Mass of the electron = $5.489\cdot10^{−4}$ atomic mass units
$\text{M}$ Mass of the nucleus
$\text{Z}$ Charge on the nucleus
$n$ orbital = principal quantum number $n$ = 1, 2, 3, 4, ...
$\text{E}_R$ Rydberg energy = 13.605693 eV

For the $1s^1$ electron of protium, deuterium and tritium, Z=1 and $n$=1.

protium
$\text{M} = 1.007825 - 0.0005489 = 1.007276$ amu

$\dfrac{1}{1+ \dfrac{\text{M}_e}{\text{M}}} = \dfrac{1}{1+ \dfrac{5.489\cdot10^{−4}}{1.007276}} = 0.999455$

ionisation enthalpy = $0.999455 \cdot 13.605693 = 13.598$ eV

deuterium
$\text{M} = 2.01410178 - 0.0005489 = 2.0135529 $ amu

$\dfrac{1}{1+ \dfrac{\text{M}_e}{\text{M}}} = \dfrac{1}{1+ \dfrac{5.489\cdot10^{−4}}{2.0135529}} = 0.999727$

ionisation enthalpy = $0.999727 \cdot 13.605693 = 13.602$ eV

tritium
$\text{M} = 3.0160492 - 0.0005489 = 3.0155003$ amu

$\dfrac{1}{1+ \dfrac{\text{M}_e}{\text{M}}} = \dfrac{1}{1+ \dfrac{5.489\cdot10^{−4}}{3.0155003}} = 0.999818$

ionisation enthalpy = $0.999818 \cdot 13.605693 = 13.603$ eV

Ionization energies From NIST webbook
hydrogen = 13.59844 eV
deuterium = 13.603 eV
tritium = 13.603 eV

Therefore protium < deuterium < tritium


I'll add that when you're doing this sort of thing other factors creep in the calculations.

(1) For a hydrogen atom the electrons aren't relativistic. If you get to uranium though that wouldn't be true.

(2) The 13.6 eV binding energy also reduces the mass of the hydrogen atom. So subtracting the rest mass of the electron isn't absolutely correct, but good enough for the precision used.

13.6 ev $=\quad\quad 0.000 000 014 5$ amu
mass e used $= 0.000 548 9$ amu

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