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Topic: QUANTUM  (Read 1910 times)

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Offline Heisenberg97

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QUANTUM
« on: April 06, 2016, 08:03:37 AM »
I've been asked to calculate the ionisation energy of hydrogen, which I've worked out at 1310 kJ/mol (to 3sf).
I now have to calculate the ionisation energy of deuterium.
I've been given the Rydberg constant (R infinity) = 109737.3 cm^-1
Since the mass of deuterium is twice the mass of hydrogen, would the ionisation energy of deuterium just be twice the value of the ionisation energy of hydrogen? (since E and m are linearly related)?

*EDIT. Or does the ionisation energy of H = ionisation energy of D, due to them having the same electrostatics? Similar to the isotope effect?

Offline Irlanur

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Re: QUANTUM
« Reply #1 on: April 06, 2016, 08:39:10 AM »
I've been asked to calculate the ionisation energy of hydrogen, which I've worked out at 1310 kJ/mol (to 3sf).

how did you calculate that? then you should be able to also calculate it for deuterium...


look up how to calculate the rydberg constant. If you calculate it with the nucleus at rest the mass doesn't matter. If you don't, you'll find the reduced mass of the electron in it, which depends on the mass of the nucleus.

Offline Heisenberg97

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Re: QUANTUM
« Reply #2 on: April 06, 2016, 11:45:54 AM »
I've been asked to calculate the ionisation energy of hydrogen, which I've worked out at 1310 kJ/mol (to 3sf).

how did you calculate that? then you should be able to also calculate it for deuterium...


look up how to calculate the rydberg constant. If you calculate it with the nucleus at rest the mass doesn't matter. If you don't, you'll find the reduced mass of the electron in it, which depends on the mass of the nucleus.


Okay I think I've got it, thank you very much for your help
I've used the formula R(H) = R(infinity)/(1+me/M)
Then v = R(H) x c
E = hv
and x Avogadro's constant to get the ionisation energy in kJ/mol

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