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### Topic: Atomic structure (Quantum, De-Broglie, Uncertainty Principle)  (Read 6148 times)

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#### Praveen27

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• Mole Snacks: +0/-0 ##### Atomic structure (Quantum, De-Broglie, Uncertainty Principle)
« on: June 13, 2015, 05:25:18 AM »

I was solving questions about Quantum Mechanical model of an atom, and came across these questions. Kindly go through them. I am not getting the proper answers according to the text. If I am completely wrong about a question, please give me the correct idea.

Q1. Calculate the accelerating potential that must be imparted to a proton beam to give it an effective wavelength of 0.005 nm.

How I proceeded: I thought of it as a matter wave, and used de-Broglie's formula.
$$λ = \frac {h} {mv}$$
$$λ =\frac{h}{\sqrt{2*m*K.E}}$$
$$λ = \frac{h}{\sqrt{2m*e*V} }$$
Using, m = 1.6 x 10-27 kg ; h = 6.62 x 10-34 Js ; e = 1.6 x 10-19 C ; and proper calculations, I found the answer to be V = 33.67 Volts. Was my approach correct? ( The answer in the text was different)

Q2. Which series is obtained in both absorption and emission spectrum?
How do I proceed?

Q3. The wave function of 3s orbital is as follows.

$$\Psi_{3s} = \frac {1}{81\sqrt{3\pi}} \left [\frac {Z} {a_o}\right]^\frac{3} {2} \left[27 - \frac{18Z}{a_o}r + \frac{2Z^2}{a_o^2}r^2\right]e^\frac{-Zr}{3a_o}$$
where r is the distance from the nucleus.

The above mentioned orbital has two nodes at 1.99 ao and x ao. Find the value of X.

I substituted r = 1.99 ao. But I got both positive values for ψ. What am I missing?

Q4. A hydrogen like atom in ground state absorbs n photons having the same energy and it emits exactly n photons when electronic transition takes place. Then the energy of the absorbed photon is given by?

Q5. The position and momentum of a 1keV electron are simultaneously determined. If its position is located to within 1 Angstrom, what is the percentage uncertainty in its momentum?
I found momentum using $$p = \sqrt{2m K.E}$$ (after proper conversion to SI units).
I found change in momentum using heisenberg's principle. I now know, both momentum and change in momentum.
I then thought that change in momentum = x% of momentum. I anyway, got a wrong answer.
Please throw me some light on these questions.

Thank you.

#### mjc123

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• Mole Snacks: +282/-12 ##### Re: Atomic structure (Quantum, De-Broglie, Uncertainty Principle)
« Reply #1 on: June 13, 2015, 06:51:34 PM »
Q1 How different is the given answer? Slightly, or orders of magnitude? Using your equation (which looks correct to me) and your numbers, I get 34.24 V. Did you use values with more significant figures? What do you mean by "proper calculation"?

Q2 What is the formula for the various series? What is constant for all absorption lines, but may vary for emission lines?

Q3 For a start, the answer depends on Z. Assuming Z=1, 1.99 is either a typo or a misreading for 1.90. Solve the quadratic equation.

Q4 I don't get this, it seems to make no sense. Did you copy it correctly?

Q5 Show your working, so we can see where you went wrong.

#### Praveen27

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• Mole Snacks: +0/-0 ##### Re: Atomic structure (Quantum, De-Broglie, Uncertainty Principle)
« Reply #2 on: June 14, 2015, 01:07:05 PM »
Q1 How different is the given answer? Slightly, or orders of magnitude? Using your equation (which looks correct to me) and your numbers, I get 34.24 V. Did you use values with more significant figures? What do you mean by "proper calculation"?

Q2 What is the formula for the various series? What is constant for all absorption lines, but may vary for emission lines?

Q3 For a start, the answer depends on Z. Assuming Z=1, 1.99 is either a typo or a misreading for 1.90. Solve the quadratic equation.

Q4 I don't get this, it seems to make no sense. Did you copy it correctly?

Q5 Show your working, so we can see where you went wrong.

Fine!
Q1. The answer given in the text was .0033 volts. 1000 times less. That's why I asked. Since you are getting an answer near mine (compared to the text, and assuming that both of us are using the correct formula), I can very well think that the text had a wrong answer! Thank you!

Q2. The formula for various series = $$\frac{1}{λ} = R_h Z^2\left[\frac{1}{n_f^2} - \frac{1}{n_i^2}\right]$$ where, Rh = 1.09 x 109 m-1

For absorption lines, ni is constant (ground state), and both nf and ni vary in emission spectrum. I know you are prompting me for the answer. It just, doesn't ring any bell. We don't get Lyman series in Absorption spectrum. (Because, an atom doesn't absorb energy and get to n = 1). Correct me if I am wrong.

Q3.So, should I substitute r = 1.9 ao? I tried doing that. But I keep getting two positive values for the wave function.

Q4. Actually, the question had options (Multi correct) with some options being a factor of 13.6. Was that supposed to ring any bell?

Q5. I want to know whether I've based my working on the correct idea? I mean, can you please state whether my approach was correct? (I mean, I found momentum; and change in momentum using Heisenberg's Uncertainty Principle. I did,  $$\Delta p = \frac{x}{100}p$$ and found the value of x. Was my approach correct?
Thank you!

#### mjc123

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• Mole Snacks: +282/-12 ##### Re: Atomic structure (Quantum, De-Broglie, Uncertainty Principle)
« Reply #3 on: June 14, 2015, 05:35:57 PM »
Q2 An absorption line going from n1 to n2 has the same energy as an emission line from n2 to n1. So series in absorption correspond to emission series, but are defined by a common initial state rather than final state. So which series do you see in absorption?

Q3 How do you get two values for the wavefunction? Inserting a value of r should only give you one. It's not exactly 1.90; solve the equation for the nodes.

Q4 Still makes no sense, except that the ionisation energy of H is 13.6 eV. But atoms don't normally absorb or emit more than 1 electron at a time. Can you quote the whole question, with the multiple choice answers, exactly please.

Q5 The approach looks correct; please show your calculation so we can see if you made a mistake or (like Q1) the book got it wrong.

#### Praveen27

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« Reply #4 on: June 15, 2015, 06:42:18 AM »
Q2 An absorption line going from n1 to n2 has the same energy as an emission line from n2 to n1. So series in absorption correspond to emission series, but are defined by a common initial state rather than final state. So which series do you see in absorption?

Q3 How do you get two values for the wavefunction? Inserting a value of r should only give you one. It's not exactly 1.90; solve the equation for the nodes.

Q4 Still makes no sense, except that the ionisation energy of H is 13.6 eV. But atoms don't normally absorb or emit more than 1 electron at a time. Can you quote the whole question, with the multiple choice answers, exactly please.

Q5 The approach looks correct; please show your calculation so we can see if you made a mistake or (like Q1) the book got it wrong.

Q2. oh YEAH! In emission spectrum, the emission lines are recorded. Lyman series is bound to come. In absorption spectrum, the dark lines are the lines of wavelengths which are absorbed. So, Lyman series cannot be dark. So, Lyman series is obtained in both Absorption Spectra and emission Spectra. Thank you! Thanks a lot, mjc123.

Q.3.You mean to say, it is not 1.99 ao , rather 1.9 alone? When I said I got 2 values for the node, I substituted x = 1.99 ao , (and the ao terms in the numerator and the denominator got cancelled) and I was left with a quadratic in r. And both values of r were irrational and positive.

Q4. A hydrogen like atom in ground state absorbs n photons having the same energy and it emits exactly n photons when electronic transition takes place. Then the energy of the absorbed photon is given by? (Multiple Correct)
a) 91.8 eV  b) 40.8 eV  c) 48.4 eV  d) 54.4 eV

Q5. Fine. Here it goes.
1keV = 1.6 x 10-16 J = K.E

So, $$p = \sqrt{2mK.E}$$
Where, m = 9.1 x 10-31 kg.

So,
$$p = \sqrt{2*9.1* 10^{-31} * 1.6 * 10^{-16}}$$
$$p = \sqrt{29.12*10^{-47}}$$
$$p = \sqrt{2.912*10^{-46}}$$
$$p = 1.706*10^{-23} m/s$$

Using Heisenberg's uncertainty principle, and assuming the minimum case:
$$\Delta x \Delta p = \frac{h}{4\pi}$$ So,
$$\Delta p = \frac{h}{4\pi\Delta x}$$
$$\Delta p = \frac{6.62*10^{-34}}{4*3.14*10^{-10}}$$
It is given that the position is located to within 1o. After simplifications, we get,
$$\Delta p = 0.527*10^{-24} m/s$$

My assumption,
$$\Delta p = \frac{x}{100}*p$$
$$x = 100\frac{\Delta p}{p}$$
$$x = 100\frac{0.527*10^{-24}}{17.06*10^{-24}}$$
$$x = \frac{52.7}{17.06}$$
$$x = 3.08$$

But the answer given in the text was 6.178. Kindly point out the mistakes (if any) in my approach.Judging by my Q.1, this book isn't much of reliable!
Thank you!

#### mjc123

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« Reply #5 on: June 15, 2015, 07:23:37 AM »
Q3 I thought you said you got two values for Ψ. Yes, you should get two irrational positive values for r/a0. One of them should be ≈ 1.90.

Q4 Sorry, I still don't get this.

Q5 Your calculation looks OK to me. The discrepancy is a factor of 2, so I think they have probably made a mistake - unless "located to within 1Å" means "located to ±0.5Å", which is not (to me) the natural interpretation of the words.

#### Praveen27

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« Reply #6 on: June 15, 2015, 08:12:13 AM »
Q3 I thought you said you got two values for Ψ. Yes, you should get two irrational positive values for r/a0. One of them should be ≈ 1.90.

Q4 Sorry, I still don't get this.

Q5 Your calculation looks OK to me. The discrepancy is a factor of 2, so I think they have probably made a mistake - unless "located to within 1Å" means "located to ±0.5Å", which is not (to me) the natural interpretation of the words.

Q.3. I still don't get it. THe value of the wave function should be zero, right? That means, at a particular distance from the nucleus ( i. e r), the wave function becomes zero. ANd isn't r = 1.9 ao and x ao ? SO, if we directly substittute r  = x ao , aren't we supposed to get the value of x directly?

Q.4. Fine. No problem. Thank you.

Q.5. So, should I go back to the conclusion of wrong answer printed in the text?

#### Praveen27

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• Mole Snacks: +0/-0 ##### Re: Atomic structure (Quantum, De-Broglie, Uncertainty Principle)
« Reply #7 on: June 15, 2015, 09:46:07 AM »
Q.5 Looks like i got the answer. I tried the way you suggested. The answer is 7. Thank you. So, Q.5 remains.

#### Enthalpy ##### Re: Atomic structure (Quantum, De-Broglie, Uncertainty Principle)
« Reply #8 on: June 16, 2015, 03:50:25 AM »
Q1. From k*5pm=2pi and E=(kh/2pi)^2/2m I too get 30eV, not meV.

Q2. Does the text specify the atom is hydrogen? If not, the absorption will start from any populated shell.

Q4. The question may perhaps suggest that emission alternates with absorption and both occur between the same pair of levels. The number of photons plays no role then.

Hydrogen-like atoms (Z protons, neutrons as needed, one electron) having electron levels Z2 times deeper than hydrogen, and still like 1/N12-1/N22, some combinations of Z, N1, N2 may fit some answers which are 6.75R, 3R, 3.559R (~32R/9), 4R.

For instance 3R is 22*(12-1/22), 6.75 is 32*(12-1/22).

#### Praveen27

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« Reply #9 on: June 16, 2015, 04:11:38 AM »
Q1. From k*5pm=2pi and E=(kh/2pi)^2/2m I too get 30eV, not meV.

Q2. Does the text specify the atom is hydrogen? If not, the absorption will start from any populated shell.

Q4. The question may perhaps suggest that emission alternates with absorption and both occur between the same pair of levels. The number of photons plays no role then.

Hydrogen-like atoms (Z protons, neutrons as needed, one electron) having electron levels Z2 times deeper than hydrogen, and still like 1/N12-1/N22, some combinations of Z, N1, N2 may fit some answers which are 6.75R, 3R, 3.559R (~32R/9), 4R.

For instance 3R is 22*(12-1/22), 6.75 is 32*(12-1/22).
Q1. Potential is measured in volts, right? Did you mean 30 V?

Q.2 It does not specify that is is Hydrogen, but yeah, Lyman series is indeed obtained in both absorption and emission series, irrespective of the element.

Q.4. Fine. But if looked at the options, we don't get a proper answer, right?

#### mjc123

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« Reply #10 on: June 16, 2015, 07:14:03 AM »
Q4 It looks as if it's saying that (where E is the photon energy)
nE = Z2*(1-1/m2)*13.6, where n, m, and Z are integers >1.
Values of Z, m and n can be found to fit any of the four answers, so I don't see the point of the question.

#### Praveen27

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« Reply #11 on: June 16, 2015, 09:42:55 AM »
OK fine. So, in Q4. the only option would be trial and error method, (i.e, finding appropriate values of Z, nf and ni.
And the doubts in all the other questions are clarified! Thank you mjc123 and Enthalpy. Thanks a lot!

#### mjc123

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« Reply #12 on: June 16, 2015, 10:34:07 AM »
Quote
i.e, finding appropriate values of Z, nf and ni.
Finding appropriate values of Z, nf and n (the number of photons). It is in the ground state, so ni is always 1.

#### Praveen27

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« Reply #13 on: June 18, 2015, 12:34:20 AM »
oh yes. How silly of me! Anyway, thanks guys!

#### Enthalpy ##### Re: Atomic structure (Quantum, De-Broglie, Uncertainty Principle)
« Reply #14 on: June 22, 2015, 07:07:46 PM »
In Q4, I don't believe that we should add photon energies. Few years ago, this would have been considered impossible; meanwhile it has been observed with very intense laser pulses, where individual photons wouldn't have enough energy to ionize nitrogen molecules, but collaborate in the strong (concentrated and ultrashort) pulses.

This must be why the exercise tells that the absorbed photons have the same energy, and that as many photons are re-emitted: to exclude successive absoprtions between varied levels, and to exclude the newer multi-photon scenario which would emit one or several photons at the atom's natural wavelengths.

Hence my understanding that photon absorptions and emissions alternate, so that the number of photons has no influence at all.

Then, at least 3*R and 6.75*R are acceptable solutions - I didn't invest more time in the other proposals.