[kyasarin]

On earth, the ionization energy of hydrogen is 1312 kJ/mol.  On a distant planet, it is so hot that all hydrogen atoms have their electron in the quantum state with n = 3.  On this distant planet, the ionization of energy of hydrogen is:
a) 437 kJ/mol
b) 146 kJ/mol
c) 1312 kJ/mol
d) 0 kJ/mol
e) none of these

I don't understand how the ionization energy of hydrogen is 1312 kJ/mol because I get 1.6 x 10-18 J when I use -2.187 x 10^-18 J (1/n_f²-1/n_i²). So i thought I would multiply it by Avogadro's number, but I got 9.6 x 10² kJ/mol.

I know that the ionization energy of hydrogen really = 1312 kJ/mol b/c I have found it online, but i only only way i can think of doing this problem is by using -2.187 x 10^-18 J (1/n_f²-1/n_i²) where n_f is 4 and n_i is 3.

I'm thinking it's possible that the answer is c) 1312 kJ/mol, but I wouldn't know why.

[Borek]

I get 1.6 x 10-18 J when I use -2.187 x 10^-18 J (1/n_f²-1/n_i²).

What values do you use for n_i and n_f - and why?

[kyasarin]

What values do you use for ni and nf - and why?

-2.187 x 10^-18 x (1/4-1) because n=1 to n=2 is the first ionization energy for H. (this is for H on earth, just to clarify)

[ARGOS++]

Dear Kyasarin;

How much less Energy is required, if “you” are already in the quantum/”exited” state 3?

Good Luck!
                    ARGOS⁺⁺

[Borek]

What is n_f for ionization? What does ionization mean in terms of electron energetic level?

[kyasarin]

the n_f = 2 on Earth and = 4 on the other planet

the n_i = 1 on Earth and = 3 on the other planet

-2.187 x 10-18 x (1/4-1) = 1.6 x 10-18 J (on earth which doesn't match the 1312 kJ/mol which is given)

-2.187 x 10-18 x (1/16-1/9) = 1.063 x 10-19 J (which doesn't match any of the answers)

I originally assumed the answer was a) 437 kJ/mol or b) 146 kJ/mol, but since i couldn't get the right answer and didn't know how i thought maybe it was a trick question and the answer was c) 1312 kJ/mol, but apparently I'm wrong.

[kyasarin]

ionization energy = the energy necessary to remove an electron from the atom

wow i just realized that ionization energy is removing an electron. 

so now i am even more lost b/c i don't think we had any formulas to find the ionization energy.

what is it called when you are "exiting" an electron - _____ energy?

wait, maybe you use n_f= ∞.

which would make it:

the n_f = ∞ on Earth and the other planet

the n_i = 1 on Earth and = 3 on the other planet

-2.187 x 10-18 x (∞-1) = ??

-2.187 x 10-18 x (∞-1/9) = ??

ok maybe not...

[kyasarin]

ok, so i think that the H atom would have the same 1st ionization energy as Na. although i find it hard to believe that they would expect us to memorize all the ionization energies, so I'm assuming the is some other way to do this problem.

the 1st ionization energy of Na = 495.8 kJ/mol which is close to a) 437 kJ/mol.

[enahs]

You are one the right track.

First, the value you are using for Rhydbergs constant is in J, but you are shooting for kJ.
Second, it is 1/∞. 1/∞ = 0

So you have
-R*(0 - 1/n²)
Plugging that in as n=1, and using the correct units, you get the energy per photon, after multiplying it by Avagrado's number, you get the correct value for here on Earth.

Note carefully that Rhydberg's constant is just the negative of the value for Hydrogen here in earth.

So now, what is it on this other planet? You do not even have to bother with Rhydbergs constant to calculate this. You know Rhydbergs constant is the Ionization energy here on Earth (which is given in the question). How does it change on another planet? (1/n²)?

[kyasarin]

-2.187 x 10^-18 x (0-1) = 2.187 x 10^-18
2.187 x 10^-18 x 6.0221415 × 10²³ = 1,317,042 J/mol
~1312 kJ/mol  thanks

So now, what is it on this other planet? You do not even have to bother with Rydberg's constant to calculate this. You know Rydberg's constant is the Ionization energy here on Earth (which is given in the question). How does it change on another planet? (1/n²)?

I don't understand why you said that. 

But...

-2.187 x 10^-18 x (0-1/9) = 2.43 x 10^-17
2.43 x 10^-17 x 6.0221415 × 10²³ = 14633803.85 J/mol
~b) 146 kJ/mol

[kyasarin]

I just wanted to thank everyone for helping me... 

[enahs]

So now, what is it on this other planet? You do not even have to bother with Rydberg's constant to calculate this. You know Rydberg's constant is the Ionization energy here on Earth (which is given in the question). How does it change on another planet? (1/n²)?

I don't understand why you said that. 

How you did it is fine.

But notice that, when n₂ = 0 and n₁=1, you are left with a -1. As you said:
-2.187 x 10^-18 x (0-1) = 2.187 x 10^-18
That is, the Rydberg constant is the ionization constant for ground state.

The only difference on the second planet is n=3. So instead of multiplying by 1, you are multiplying by 1/9th (dividing by 9). That is saying that the ionization energy is 1/9th. Yes you can go through and calculate it, or you can just take the ionization given (1312) and divide by 9!

That is all I was trying to say.