Iron is commonly found as Fe, Fe2+, and Fe3+.
(c) The third ioniz...

Question

Iron is commonly found as Fe, Fe2+, and Fe3+.
(c) The third ionization energy of Fe is Ei3 = +2952 kJ/mol. What is the longest wavelength of light that could ionize Fe2+1g2 to Fe3+1g2?

Accepted Answer

Welcome back everyone. We're told that mercury exists in nature as H G H. D plus one and H. D. Two plus if mercury has a second ionization energy of positive 1810 kg per mole, calculate the longest possible wavelength in nanometers of light that can remove an electron from a mercury plus one can of gas to a Mercury two plus category of gas. So our first step is to recall our formula which relates the energy of a photon of light to plank's constant times the speed of light. C divided by lambda, represented by our wavelength. Or sorry, our wavelength represented by lambda. And we're going to reference this formula so that we can solve for our energy using the given ionization energies where from the prompt were given a second ionization energy for mercury, H G equal to plus 1810 kg joules per mole. And so relating this to our formula, we can say that the energy of our photons needed to remove an electron, which we want to recall is represented in units of joules per photon is going to be equal to that ionization energy given in the prompt being 1810 kg joules per mole. Where we want to convert from killing joules per mole again to jules per photon. So first focusing on getting rid of killer jewels, two jewels, we want to plug in kila jewels and jewels in our numerator were killed jules is in the denominator and our prefix kilo, we would recall, tells us that we have 10 to the third power of our base unit jewels. So canceling out kill jules. We're now going to introduce our unit of photons using avocados number where we would recall that it tells us that we have six point oh 22 times 10 to the 23rd power photons with an equivalent to one mole. And so canceling out our units of moles were left with joules per photon as our final unit. And this is going to result in an energy of 3.56 times 10 to the negative 18th powered jewels per photon of energy required to remove the electrons. But we need to figure out the wavelength required to remove the electron from the mercury plus one catty on to the mercury. Two plus Catalan. So we're going to now recall a second formula where, sorry, not a second formula, but a second arrangement of our formula. Where we are isolating for wavelength or lambda. And setting that equal to Planck's constant. Times our speed of light, divided by E. For our energy of our photon. And so plugging in what we know, we have planks constant, which we should recall from lecture is 6.626 times 10 to the negative 34th power jewels, times seconds multiplied by our speed of light, which we recall from lecture is three point oh oh times 10 to the eighth. Power meters per second. And then divided by our energy of our photon, which we just calculated above. Being equal to six point or sorry, being equal to what we calculated as 3.56 times to the negative 18th power jewels per photon. And so canceling out our units. We can get rid of jewels as well as seconds. And because we understand that photons was just a derived unit from avocados number. We're also going to cancel that out, leaving us with meters as our final unit. And this is going to yield a wavelength of 6.61355 times 10 to the negative eighth power meters. However, as the prompt states, we should report this wavelength in nanometers. So we're going to convert from meters into nanometers by recalling that our prefix nano tells us that we have 10 to the negative nine power and we should have 10 to the negative ninth power of our base unit meters equivalent to one nanometer. So this allows us to cancel out meters, leaving us with nanometers as our final unit of wavelength, which is going to result in a wavelength equal to 66.14 nanometers. So this would be our final answer as our wavelength. That is the longest possible wavelength of light that can remove an electron from a mercury plus one cat I own of gas to a to form a mercury two plus Catalan of gas. So it's highlighted in yellow is our final answer. I hope everything I explained was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.

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