What is the Ionisation Energy of Hydrogen in KJ mol-1?

[crays]

Hi, I've a question that shows me a diagram for wavelength, there's 4 of them

11.1 x 10⁶
10.5 x 10⁶
9.7 x 10⁶
8.4 x 10⁶

then it asked me to determine the ionisation energy of hydrogen in KJ mol^-1 by using the above spectrum.

From what i know
E = hf
f = c/lambda
lambda being the wavelength

The equation should be using the largest frequency so i should pick the shortest wavelength right? which is 8.4 x 10⁶

But the calculation behind uses the value of 11.1 x 10⁶ . Can anyone explain?

 

[drizzle]

You know that the electron of a hydrogen atom has 13.6 eV less energy than a motionless electron infinitely far from the nucleus. Thus, the http://en.wikipedia.org/wiki/Ionization_energy" needed to set this electron free is supposed to equal 13.6 eV or more but not less. So, among the wavelengths you've stated, which one would provide the sufficient energy needed to set it free?

 

[chemisttree]

What are the units for wavelength?

 

[drizzle]

11.1 x 10⁶
10.5 x 10⁶
9.7 x 10⁶
8.4 x 10⁶

I think these numbers are to the power of -6!

 

[Redbelly98]

What are the units for wavelength?

Good question. crays, can you tell us what units are used for wavelength in the problem statement.

I think these numbers are to the power of -6!

This does not answer the question about units, but if it's true than we need crays to clarify whether the exponents are +6 or -6.

 

[drizzle]

This does not answer the question about units, but if it's true than we need crays to clarify whether the exponents are +6 or -6.

I wasn't answering chemisttree's Q, but I think it's meters in this case, probably.

 

[crays]

So sorry for the late reply, it is written

wave number, v (x 10⁶ m^-1) by m^-1 i assume it is wave length.

From my book it says that frequency is equals to c/lambda but not proportional to. Clarification please. cause if its wavelength, it should be h x 1/wavelength x c.

 

[Redbelly98]

Wave number is 1/λ, i.e. the reciprocal of the wavelength. That is why the units are m^-1 instead of m.

The equation should be using the largest frequency so i should pick the shortest wavelength right? which is 8.4 x 10⁶

But the calculation behind uses the value of 11.1 x 10⁶ . Can anyone explain?

The frequency is simply c·(1/λ), or c-times-wavenumber. Hence the largest frequency goes with the largest wavenumber, 11.1 x 10⁶ m^-1.

EDIT:

From my book it says that frequency is equals to c/lambda but not proportional to. Clarification please. cause if its wavelength, it should be h x 1/wavelength x c.

The frequency is c/λ. The photon energy is hc/λ = h x frequency.

If you examine the units in h, c, and λ, you'll find the expressions in the above paragraph work out to s^-1 for frequency and J for energy, just as they should.

 

[crays]

Thanks a lot for clarification and the help :)