5

(c) 4, 3, 0,3,2, -2, 2Is the transition allowed?Select--_What is the energy involved (in eV)? (Include sign. Enter "none" if the transition is forbidden:)...

Question

(c) 4, 3, 0,3,2, -2, 2Is the transition allowed?Select--_What is the energy involved (in eV)? (Include sign. Enter "none" if the transition is forbidden:) evIs the photon absorbed or emitted?Select--(d)2, 1, 0, ~2) - (4,0,-3, Is the transition allowed? ~~Select---What is the energy involved (in eV)? (Include sign. Enter "none" if the transition is forbidden.) evIs the photon absorbed or emitted? Select--

(c) 4, 3, 0, 3,2, -2, 2 Is the transition allowed? Select--_ What is the energy involved (in eV)? (Include sign. Enter "none" if the transition is forbidden:) ev Is the photon absorbed or emitted? Select-- (d) 2, 1, 0, ~2) - (4,0,-3, Is the transition allowed? ~~Select--- What is the energy involved (in eV)? (Include sign. Enter "none" if the transition is forbidden.) ev Is the photon absorbed or emitted? Select--



Answers

Which of the following transitions are allowed for an electron in hydrogen? For each allowed transition, find the wave- ; (b) $3 d$ to $2 s$; (c) 4 fto length of the emitted photon: (a) $3 p$ to $2 s$; $3 d ;$ (d) $3 p$ to $1 s$

So we're given a set off atomic transitions here. We've got transitions. A T E. These are all emissions, by the way, because they're going to Ah, lower state. And so we asked our views. Transitions allowed or forbidden. So the selection a little but we need to remember is that our delta l, which is the, um, angular momentum quantum number can be plus or minus one. Um, it can't be anything else. Otherwise, the transition is forbidden. And we know that the angle momentum, quantum number l corresponds to whatever orbital the sub state is in. So we have l equals 0123 here. And we know that this is equivalent to the S p d f on etcetera, etcetera, sub shells that we are in. So, using that information, we can deduce whether or not these transitions are allowed or forbidden. So the four p to the three p um, transition. This is forbidden on The reason it is forbidden is because our we're going from a P orbital to a P orbital on this corresponds to a delta l off. Well, zero andi. Therefore, it's not allowed because we need plus or minus one so I just didn't know it there. That's not equal to plus or minus one. The second transition, this is allowed because, as you can see, our delta l is equal to minus one here, which is equal to obviously plus or minus one. We're going from a P orbital to an s orbital. That's an l of one going to knell of zero. Uh, see, um, this one is forbidden. You should be able to see this by now on. The reason is because we are going from a Delta l off. Well, what's are changing? L We're going from a deal. Little to a deal. Battles that zero, which is obviously not plus and minus one, which is what we want. Um, de um this one is also forbidden on The reason for this is because we're going from an l, um, corresponding to the deal brittle, which is an l of to to an l off zero. So a delta l is minus two, and this is obviously not equal to plus or minus one. Andi, finally, e. This transition is from a P orbital to an s orbital. So we have an l off. Sorry. Inaccessible to to appeal, but also we've got l off zero going to Nell off one. So this is allowed on. The reason it is allowed is because we are going from en El off zero to Anel of one. So this Adele's l off plus one. So these are answers. If you can remember the selection rule, it's pretty straightforward.

Are the only transitions allowed. Uh, are those that have ah, changing principal quantum number. And, uh, which is odd. Okay, I'll changes and dealt and on. So, um, energy level diagram looks like Yes, this is equals four equals three and equals two on an equals one. You could have odd number transition. So from here, you can see the delta and can only be one or three. All the jumps, um, in of one energy level are good. So pinnacles for two and equals 43 3 to 22 to one. All good, but also Ah, this one jump for men equals or two n equals. One is also allowed. So those are the four allowed transitions for part B. We note that energy of an infinite square. Well, uh, in the level is given my end. Square times are X squared over eight ml squared. So this is and squared times. Plank's constant. 61 66 has turned a negative. 34 jewel seconds over electron mass 9.11 times 10 to the negative. 31. Um, kilograms looks, kilograms times 0.2 nanometers. So times tending negative line meters squared. Um And so once you convert this to from Jules to electron volts by dividing by 1.6. So two times 10 a negative 19. Ah, you get and square times 9.4 for these therefore change. Uh, the change in energy level when you're going from n equals of when you're going from and I an initial energy level 10 have final energy level. That's 1/4 uh, and I squared minus and squared. Times 9.4 electron volts canceled. Let's, um let's work those out. Do you have some say so. Delta, you have, um, does it give of, um, and going from 4 to 1? So this will be, ah, four squared minus one times 9.4 is equal to 1 41 electron volts, and you have four. To me, this is four squared minus to B squared times nine point floor. Just a quote to 65.8 electron volts. Then you have change from ah equals to meet it too. So that will be three squared, minus two squared times nine point floor. That's 47 electron volts finally dealt it. He can go from 2 to 1 2 to 1. So that gives you two squared minus one times 9.4 uh, equals 28.2 electron volts. And so these are all And so these a rolled up impossible, uh, transitions.

So we know that traditions are only possible. If and got him is equal. Go less minus one. So for the first part with the condition and dockers from four p, the creeping does that end is equal to zero because l is one in boat. So this transition it's forbidden for the second part where there is a tree be two one this transition that will be one minus Edo which is equal to one. Yeah, so this is equal to one. So this transition is allowed at the time fied forty Jody Delta Alice against zero. So this transition is forbidden. Forty five day two three s Delta is's to minus legal. This is for real. Then again, And similarly for the last part, we have the four less too topi transition. So don't I will be minus one. He'LL smile minus one zero, minus one. So that's minus one And this is allowed

In this exercise, we have four transitions of the hydrogen atom where the Adam transitions from an initial energy level and more and I two final energy level and f ah, And in order to calculate to answer all the questions that are asked of us here, the first thing I'm going to do is to calculate Alto E. That is the difference in energy between the final energy GF and the initial energy e I and I have to remember here that the energy of I'm gonna write it in blue here that the energy off the end energy level of the hydrogen atom is given by minus 13.6 electoral votes divided by and square. So this means that Delta E, which is e f minus, equals B F minus E. I is just minor is just 13.6 times one over, and I square mine is one over in F square. Okay, so I can calculate the difference in energy for each of the four transitions you. So let's do it first. The difference in energy for transition one that's gonna be 13.6 times one over an ice square, and I is too So it's one of her four minus one over an F squared, which is one over 25 and this is equal to 2.85 six electoral votes. Then we have the difference in the energy for the transition to, and that's 13.6 times and I, which is and I squared, which is 1/5 square, that is, 1/25 miners one over and naff squared, which is 1/9. And this is minus zero point 967 Noticed that here feel the Adam loses energy, and in this process it emits ah, foot. For the third doubt, E three, we have 13.6 times, 1/49 which is 1/7 square. Mine is one over 16. This is minus zero point 572 electoral votes. Finally, for the fourth, we have 13 point 6/1 times 1/16 minus 1/9. 49 and this is zero point 572 electoral votes. Okay, okay, so now we are in a position to answer the questions in question. A. We asked what is the transition that results in the short, the shortest wavelength off the folding being emitted. So we have only two transitions here that Emmett Fulton's okay, we have the and the transitions that emit Fulton's are all always have a negative difference in energy. Okay, so the elect, the Adam loses energy, and that's why it can emit a phone. So we had of the transition to in the transition three. Okay. And we're looking among them. Which one has the shortest wavelength? Lunda. And we know that the wavelength of the photon is given by HC over the energy of the photo. Okay, And remember that the energy of the Fulton is minus the difference in energy of the ah, of the atom. So the biggest the energy of the election. Okay, the smallest, the the energy. I'm sorry. The smallest, the wavelength of the photon. So we're looking core the difference in energy of the atom that has the the greatest magnitude. Okay, because then the the energy of the fourth movie, Greatest and then the wavelength of the fulton will be the shortest. And that happens for transition to because the magnitude of the surface of the difference in energy into the transition to 0.9 67 electoral votes, and that's greater than the magnitude off 0.572 So the maximum the actually the shortest wavelength Lambda is that happens for transition to the question be we have to say wish transition. In which tradition that the Adam gains the most energy. So we have to see which transitions are positive and that's just one and four. And we have to see among them which one has the greatest energy. And obviously it's the 1st 1 It's 2.80 uh, 856 So the answer is transition to and in C were asked for wish musicians. The Adam loses energy, and that's all we have to do. Still, look at the differences in energy and see which ones are negative. So we have the two and three are negative. Uh, yeah, I said that actually, India had two. It's one right. I pointed to the right one, but I wrote that there are and in seeds two and three. Okay. And this concludes that the exercise


Similar Solved Questions

5 answers
7. Which of the following can be determined from the graph below? The ordered pair (6,25) is a solution to bothy = Ex 40 andy 5x+ B. The ordered pair (25,6) is solution to bothy 5+ 40 andy %+ C. The ordered pair (6, 25) is a solution to bothy = 5x 40 andy = D. The ordered pair (6, 25) is a solution to bothy- #+ 40 andy=3*+15
7. Which of the following can be determined from the graph below? The ordered pair (6,25) is a solution to bothy = Ex 40 andy 5x+ B. The ordered pair (25,6) is solution to bothy 5+ 40 andy %+ C. The ordered pair (6, 25) is a solution to bothy = 5x 40 andy = D. The ordered pair (6, 25) is a solution ...
5 answers
25 2+ has how many electrons? SsMn16.
25 2+ has how many electrons? SsMn 16....
5 answers
JoionnsuMIELE'bujutewej sidwbue paliuiiun JAEunOX 53un wajqold 5iy1 paidwaile JAEY nOAsjovasuy uwgnsSja"SUY Malxajd35im*D0nj-unon poiubio DuE Uibuo 041 jE pajeiuej MOILInJJO Jum Jieinjiez 40J0041 5 4bbjs Osn (fre- Jiz "I) osoddns (1d 4)snipei j0 0 8pipjJaleimneddunoupJXON^aldwaiqold :WaJOa4L suaaJI 61 MHMojogui 5u0ou6 61 m4 uejienojo}- Lloziie}-szozyew QomqomYjia]ny J0 NOLLYIJOSSY IVJILVNJHIYM MHW
JoionnsuMIELE 'bujutewej sidwbue paliuiiun JAEunOX 53un wajqold 5iy1 paidwaile JAEY nOA sjovasuy uwgns Sja"SUY Malxajd 35im*D0nj-unon poiubio DuE Uibuo 041 jE pajeiuej MOILInJJO Jum Jieinjiez 40J0041 5 4bbjs Osn (fre- Jiz "I) osoddns (1d 4) snipei j0 0 8pipj Jaleimned dunoup JXON ^ald...
3 answers
In each of these problems a set is given and a relation on the set is de- scribed . Fill iu the table below to indicate which of the properties (rellexive; symmetric; antisymmetric, transitive; equivaleuce relation , partial orderiug elation) are possessed By the relations in the exercises When the relation has the property; usC: a check mark indicate this_ (30 points)For subsets S and T of {1,2,3. five elements10}, SpT means S' and T both have
In each of these problems a set is given and a relation on the set is de- scribed . Fill iu the table below to indicate which of the properties (rellexive; symmetric; antisymmetric, transitive; equivaleuce relation , partial orderiug elation) are possessed By the relations in the exercises When the ...
5 answers
Determine whether the two triangles can be proven congruent using the ASA congruence method: If so, select the congruence statement:A) The triangles aren't congruent using ASA.B) AEFG = AHGF0Oc AEFG = AHFGD) AGFE = AFGH
Determine whether the two triangles can be proven congruent using the ASA congruence method: If so, select the congruence statement: A) The triangles aren't congruent using ASA. B) AEFG = AHGF 0Oc AEFG = AHFG D) AGFE = AFGH...
5 answers
Company estimates that the value of its new production equipment depreciates at the rate ofdv = 5000 (t _ 11) where 0 < t < 11 dtwhere V gives the value of the equipment after t years. Find the total decline in value of the equipment over the first 4 years .a)$302,500b)S35,000$122,500S180, 000$20,000None of the above
company estimates that the value of its new production equipment depreciates at the rate of dv = 5000 (t _ 11) where 0 < t < 11 dt where V gives the value of the equipment after t years. Find the total decline in value of the equipment over the first 4 years . a) $302,500 b) S35,000 $122,500 S...
5 answers
Describe in words and sketch the regids, in R? represented by the equations:(a)X=y 2 = 0(b) Y=*
Describe in words and sketch the regids, in R? represented by the equations: (a) X=y 2 = 0 (b) Y=*...
5 answers
21.30worksheets ~all subjects-pdfDasic Potahoas i^ R3 Pola 6Jn Px(e) (48 51n& 010u4 ( ~axi: sine 6056Cose-SingRy (0)Sxne(01B6016se2z (0)c08509Tnuesdgate X repe hes o (0 ta hos R : Jnuefs0 s oboue Eigeuelus ? Etqouckrs Alarlo' j Jjo Poaaleabl 6) Write aboue Yofahons Br =I, &=f) 8 = I 8 = T
21.30 worksheets ~all subjects-pdf Dasic Potahoas i^ R3 Pola 6Jn Px(e) (48 51n& 010u4 ( ~axi: sine 6056 Cose -Sing Ry (0) Sxne (01B 6016 se 2z (0) c08 509 Tnuesdgate X repe hes o (0 ta hos R : Jnuefs0 s oboue Eigeuelus ? Etqouckrs Alarlo' j Jjo Poaaleabl 6) Write aboue Yofahons Br =I, &...
5 answers
Solve_ 3+30 € = -160 € =+30 : =160 € = 2
Solve_ 3+3 0 € = -16 0 € =+3 0 : =16 0 € = 2...
5 answers
Compute the real part; the imaginary part, and absolute number of the complex number 2 + j a = 1 +2j
Compute the real part; the imaginary part, and absolute number of the complex number 2 + j a = 1 +2j...
5 answers
13. [-/1 Points]DETAILSLARCALCIT 5.1.092.MI:Locate the relatlvc extremum and point of inflection. Use graphing utillty to confirm your results_ Y = Ix2 'In(4)relative extremumpoint of InflectionNeed Help?Submit Assignment Save Assignm
13. [-/1 Points] DETAILS LARCALCIT 5.1.092.MI: Locate the relatlvc extremum and point of inflection. Use graphing utillty to confirm your results_ Y = Ix2 'In(4) relative extremum point of Inflection Need Help? Submit Assignment Save Assignm...
5 answers
Sren tv iisinotix n 0 Z1 > Y End 24- 4>8 Vse m Gnotng Inal (o pteph ia Inierenctancrck t0 etUEs BE
Sren tv iisinotix n 0 Z1 > Y End 24- 4>8 Vse m Gnotng Inal (o pteph ia Inierenctan crck t0 etUEs BE...
5 answers
Problem 6 (10 points). A sample of natural gas contains 12 moles of methane (CH;), 2.0 moles of ethane (CzHe) , and 6.0 moles of propane (C,H;) What is the partial pressure of ethane, if the total pressure of the mixture is 2.0 atm? Enter your answer In Ihe box provided with correct unils and sig filgs. Prok The partial pressure Answer: of ethane IS Pczhs
Problem 6 (10 points). A sample of natural gas contains 12 moles of methane (CH;), 2.0 moles of ethane (CzHe) , and 6.0 moles of propane (C,H;) What is the partial pressure of ethane, if the total pressure of the mixture is 2.0 atm? Enter your answer In Ihe box provided with correct unils and sig fi...
5 answers
Is the major product for the following alkylation Is it performed under kinetic = thermodynamic conditions?Lo4/ 7e %EQuestlon 9Rank the following compounds in order of reactivity in nucleophilic addition from mostto eastB>C >0 >4D>B>A>CD>C >B > 4B>D>C >4
is the major product for the following alkylation Is it performed under kinetic = thermodynamic conditions? Lo4/ 7e %E Questlon 9 Rank the following compounds in order of reactivity in nucleophilic addition from mostto east B>C >0 >4 D>B>A>C D>C >B > 4 B>D>C >4...
5 answers
A total resistance of 3.34 Q is to be produced by connecting an unknown resistance to a 12.07 0 resistance: (a) What must be the value of the unknown resistance; and (b) should it be connected in series or in parallel?NumberUnits(b)in parallcl
A total resistance of 3.34 Q is to be produced by connecting an unknown resistance to a 12.07 0 resistance: (a) What must be the value of the unknown resistance; and (b) should it be connected in series or in parallel? Number Units (b) in parallcl...
5 answers
Which sequence of transformations will map figure H onto figure H'?17 >07 ^ 5 6 7 8 9 " 72 H'
Which sequence of transformations will map figure H onto figure H'? 1 7 >0 7 ^ 5 6 7 8 9 " 72 H'...
5 answers
VtW In nuftoniel Incane lomtuch (t7"UorcumJietRcanc#
VtW In nuftoniel Incane lomtuch (t7" UorcumJiet Rcanc#...

-- 0.018003--