Consider the electronic structure of the element bismuth.
(a) The...

Question

Consider the electronic structure of the element bismuth.
(a) The first ionization energy of bismuth is Ei1 = +703 kJ/ mol. What is the longest possible wavelength of light that could ionize an atom of bismuth?

Accepted Answer

Welcome back everyone. We're told that potassium has a first ionization of plus 4 18.8 kg per mole, determine the longest possible wavelength in nanometers of light that can ionized a potassium atom. So we want to recall our formula which relates the energy of a photon of light equal to Planck's constant H. Multiplied by the speed of light C divided by lambda. Our wavelength where we want to recall that our energy of our photon should be expressed in units of joules per photon. So we're gonna need to utilize that ionization energy for potassium given it a prompt as 18.8 kg joules per mole. However, we need to convert this to jules per photon by first focusing on canceling out killer jewels and converting to jules. So we're going to recall that our prefix kilo tells us that we have 10 to the third power of our base unit jewels. So now canceling out killer jewels, We want to introduce our units of photons and cancel out moles by using avocados number, which tells us that we have six point oh 22 times 10 to the 23rd power photons with an equivalent of one mole. So canceling out moles were left with joules per photon as our final unit for our energy And this is going to yield a energy of 6.9545, Sorry 6.9545 times 10 to the negative 19th power jewels per photon. So now that we have this ionization energy now reported in jewels per photon were now going to rearrange our formula to solve for wavelength lambda. So we would have wavelength equal to our planks constant, multiplied by the speed of light but now divided by our energy. And so plugging in what we know, we have our value for planks constant, which we recall from lecture is 6.626 times 10 to the negative 34th power with units of 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. Which is then divided by our energy that we just calculated to be 6.9545 times 10 to the negative 19th power joules per photon. And so canceling out our units, we can get rid of jewels with jewels and denominator as well as seconds with seconds here. And then because we understand that photon is are derived unit using avocados number, we're going to just cancel it out, meaning that we're just left really with meters as our final unit, which is what we want for our wavelength so far. And so plugging everything carefully into our calculators were going to yield a wavelength of 2.858 times 10 to the negative seven power meters, which we now want to focus on converting into nanometers because the prompt states to give the wavelength in nanometers. And so we're going to recall that our prefix nano tells us that we have 10 to the negative nine power of our base unit meter equivalent to one nanometer. So canceling out meters, we're left with nanometers now. And this is going to yield our wavelength of 85.8 nanometers. And so for our final answer, we have our wavelength specifically our longest possible wavelength in nanometers of light that will ionized our potassium atom. So what's highlighted in yellow is our final answer. I hope that everything I reviewed 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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