What is the wavelength λ of the photon that has been released in ...

Question

What is the wavelength λ of the photon that has been released in part b?

Accepted Answer

Welcome back everyone. We need to determine the wavelength of light emitted when an electron in a hydrogen atom transitions from the third energy level. N equals three to the first energy level and equals one. Where does the emitted light fall on the electromagnetic spectrum? So we're going to begin by recalling our bomber equation in which we would take our inverse wavelength and relate that to our Rydberg constant represented as are multiplied by the atomic number of hydrogen represented as Z squared, which is then multiplied by the difference between the falling quotient. Where we have one over our final energy level squared subtracted from one over our initial energy level squared. And there should be parentheses here in the denominator. And just to be clear, we have lambda representing our wavelength. Now recall that our units for wavelength are going to be in meters. Recall that our our Rydberg constant represents a value of 1.97 times 10 to the seventh power inverse meters. And according to the prompt, we are dealing with the atom hydrogen. And so recall that hydrogen has an atomic number equal to one based on its position in the periodic table Where we can see that it has atomic number one above it in group one a according to the prompt, our final energy level is going to be and equals one because that is where our electron lance. And then our initial energy level is going to be where the prompt mentions our electron takes off from which is that unequal to three. And so now that we have everything outlined, we should plug in everything into our formula that we know. So we're going to be solving for the wavelength. So we have our inverse wavelength so far set equal to our Rydberg constant, which we will plug in as 1.97 times 10 to the seventh power inverse meters, plugging in our atomic number of hydrogen, which we agreed is one which is squared. And then in our brackets, we have our value of one over our final energy level being one squared. And then this is subtracted from one over our initial energy level squared, which is 1/3 squared. So finishing off our brackets, we will begin by taking the product Of these two terms first. Or actually rather we would begin with our brackets first. So we have one over lambda. And sorry, this is lambda here set equal to we can take the products and the bracket step at the same time. So the product is going to result in a value as 1.97 times 10 to the seventh power. We still have units of inverse meters when we multiply by one squared. And then in our brackets, which will no longer be brackets will just have parentheses which from the results of our brackets, we have a value of equal to a value of zero 0.888889. And then now taking the product of these two numbers on the right hand side, we have our left hand side. One over lambda equal to the right hand side which results in the product as 9.75111 times 10 to the sixth power inverse meters. Recall that when we have diagonals in algebra we can just switch places. And so that would give us now one or rather lambda isolated and equal to 1/9 0.75111 times 10 to the six power inverse meters, in which the result of the quotient will allow us to find our wavelength equal to a value of 1.2552. And sorry, let's rewrite that 1.0255, 2 Times 10 to the -7 power. And we no longer earn a fraction. So we have regular units of meters. And on our electromagnetic spectrum this will correspond to the ultraviolet region based on that wavelength that we calculated. And so for our final answer, highlighted here in yellow, let's actually round this wavelength to our minimum number of sig figs being three Sig figs. So we would have 1.3 times 10 to the negative seventh power meters. So what's highlighted in yellow is our wavelength of light that is emitted when an electron for a hydrogen atom transitions from the third energy level and lands at the first energy level. And for the second part of our answer, we have that this wavelength corresponds to light on the ultraviolet region of our electromagnetic spectrum. And so these answers will correspond to choice be in the multiple choice. So I hope this was helpful and let us know if you have any questions.

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What is the wavelength λ of the photon that has been released in part b?

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