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Offline jayjay112

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Ionisation Energy
« on: April 06, 2009, 02:28:08 PM »
The Lyman spectrum (n1 = 1) spectrum of Li2+ contains consecutive lines with wavelengths of 13.5 nm and 11.4 nm. What is the ionization energy of Li2+?

Can somebody please show me the method of answering this question.
Like what formulas to use??

I have an answer here http://www.slideshare.net/Mrreynon/chapter-30-powerpoint-presentation
on page 11 of 13

Thanks.

Offline cliverlong

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Re: Ionisation Energy
« Reply #1 on: April 06, 2009, 05:12:22 PM »
The Lyman spectrum (n1 = 1) spectrum of Li2+ contains consecutive lines with wavelengths of 13.5 nm and 11.4 nm. What is the ionization energy of Li2+?

Can somebody please show me the method of answering this question.
Like what formulas to use??

I have an answer here http://www.slideshare.net/Mrreynon/chapter-30-powerpoint-presentation
on page 11 of 13

Thanks.
The following might work

Slide 30.2 gives the reciprocal of the wavelength for the wavelength of light emitted when electron falls from energy level n to 1. You have two wavelengths so you should be able to determine a value for R by elimination between two equations.

Now, what energy level (n)  is the electron at when it is ionised?

What does that mean for the value 1/n2 ?

So you can now deduce a value for λ when Li2+ is ionized by electron loss.

Slide 30.3 gives a expression for energy change in terms of Planck's constant (h) and the frequency (f).

Do you know an expression to relate λ to f?

Can you now calculate the energy change for this transition (the ionisation energy)?

Clive

Offline Borek

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Re: Ionisation Energy
« Reply #2 on: April 06, 2009, 05:42:05 PM »
It is not very different from hydrogen atom...
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Offline jayjay112

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Re: Ionisation Energy
« Reply #3 on: April 07, 2009, 05:53:09 AM »
so i use this formula 1/landa = R ( (1/n1^2) - (1/n2^2))

or this E = hRD((1/n1^2) - (1/n2^2))  where hRD is the ionisation energy.

Do i let n1 = 1? why?
What do i let n2 = ? why??

Offline Borek

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Re: Ionisation Energy
« Reply #4 on: April 07, 2009, 08:26:04 AM »
So you are using formula not understanding what symbols mean? That won't get you far.

What is n1? What is n2?
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Offline jayjay112

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Re: Ionisation Energy
« Reply #5 on: April 07, 2009, 09:12:02 AM »
n1 is the first enegry level so does that mean n2 =2?

i know what n is i just dont understand why n has different values and when they should be different.

Offline Borek

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Re: Ionisation Energy
« Reply #6 on: April 07, 2009, 10:34:21 AM »
i know what n is i just dont understand why n has different values and when they should be different.

Sorry, but you are contradicting yourself - if you know what are n1 and n2, you should know why they have to be different and what it means...

What is n1? What is n2?

What are these formulas:

1/landa = R ( (1/n1^2) - (1/n2^2))

or this E = hRD((1/n1^2) - (1/n2^2))

Are they different, or not?
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Offline lancenti

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Re: Ionisation Energy
« Reply #7 on: April 07, 2009, 11:56:48 AM »
n1 is the first enegry level so does that mean n2 =2?

n1 is the first energy level, and n2 is the second energy level.

i know what n is i just dont understand why n has different values and when they should be different.

Well, think of it this way.

The second energy level, is not the same as the first, otherwise they wouldn't call it the second energy level, right?

If that's the case, it has to have either a higher or lower energy, right?

Since we're talking about n1, which is the lowest energy state of an electron in an atom, n2 has to be higher in energy, right?

Offline jayjay112

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Re: Ionisation Energy
« Reply #8 on: April 07, 2009, 12:21:21 PM »
thanks, so if n1 is 1 then n2 is 2.

So if i use the E=hRD formula twice, once for each wavelength and then get the average of these
values that is the method of answering these types of questions???

Thanks for the help

Offline lancenti

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Re: Ionisation Energy
« Reply #9 on: April 07, 2009, 12:43:41 PM »
The problem with this question is that they just tell you that the two lines are adjacent. This means that it's not necessarily n1 and n2 but na and na+1.

The method presented in the slides appears to be assuming that the Lithium ion is like a hydrogen atom already and honestly don't see how the wavelengths given in the question are useful using that method.

You can use the wavelengths to find out Rydberg's Constant, R. Once you've done that, you can find the limit of lambda as n goes to infinity, as per cliverlong's suggestion. Then convert that lambda into a frequency, then using the energy equation you have, you should get the ionization energy you need.

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