Calculate the physical properties of a Bohr atom

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Homework Help Overview

The discussion revolves around calculating the physical properties of a Bohr atom, specifically focusing on the carbon ion C^5+. Participants are tasked with determining the radius of the n=1 orbit, the energy of the electron in that orbit, and the wavelength of radiation emitted during the Lyman alpha transition.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants explore the use of the Bohr model equations to find the radius and energy of the electron. There are discussions about the correct atomic number for carbon and the implications of C^5+ as a one-electron system. Questions arise regarding the identification of transition states in the Lyman series and the validity of calculations based on different assumptions.

Discussion Status

The discussion is active with various interpretations being explored. Some participants provide calculations and reasoning, while others question the assumptions made, particularly regarding the atomic number and the transitions involved in the Lyman alpha series. There is no explicit consensus on the correctness of the calculations, but guidance is offered on the application of the Bohr model.

Contextual Notes

Participants note confusion regarding the atomic number of carbon and the implications of C^5+ in the context of the Bohr model. There is also a recognition that the Lyman alpha transition specifically refers to a transition from n=2 to n=1, which is a point of clarification in the discussion.

Benzoate
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Homework Statement



What is the radius of the n=1 orbit in C^5+ ? What is the energy of the electron in that orbit? What is the wavelength of the radiation emitted by C^5+ in the lyman alpha transition

Homework Equations



1/lambda=R*(1/(n^2)-1/(m^2) ); a(0)=(h/(2*pi))/(m*k*e^2);a(0)=.0529e-9m =5.29e-7 mE=-k*Z^2*e^2/2/((r)); r= a(0)*n^2/Z

The Attempt at a Solution


to find the radius , I know r=a(0)*n^2/Z ; Z is the atomic number. The atomic number of C^+5 is 5. r=(5.29e-7 m)*(1)^2/(5)= 1.06e-7 m

to Find energy E=-k*Z^2*e^2/(2*r)= (9e9)((5)^2)((1.609e-19 J)^2)/(2*(1.06e-7m))=2.74e-20=.170 eV

Now I'm asked to find the wavelength in the lyman alpha transition. Since I'm in the lyman transition , should I say one the transition state n=1? Maybe I don't need to find the other transition state. I could used the fact that 1/lambda=(1/hc)*(E(f)-E(i)) . I know from earlier in the problem that E(i) = .170 eV. For a lyman series E(f)= 13.6 eV . E(f)-E(i)/(hc)=1/lambda => lambda= hc/(E(f)-E(i))
 
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The Lyman alpha transition is from m = 2... intial orbit, to n = 1... final orbit...

so substitute these values into the formula and you'll get the wavelength.
 
how do you know one of the transitions is m=2?
 
so the rest of my calculations are okay? . USing the fact that E(f)-E(i)/(hc)=1/lambda is an option for finding lambda to right since E(f)=13.6 eV for the lyman series
 
Benzoate said:
so the rest of my calculations are okay? . USing the fact that E(f)-E(i)/(hc)=1/lambda is an option for finding lambda to right since E(f)=13.6 eV for the lyman series

I'm not sure... what does C^5+ mean?
 
learningphysics said:
I'm not sure... what does C^5+ mean?

C^5+ is Carbon and it reads that carbon has an atomic number Z +5. I finding some confusion here: according to the periodic table, the atomic number of carbon is 6.
 
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Benzoate said:
so the rest of my calculations are okay? . USing the fact that E(f)-E(i)/(hc)=1/lambda is an option for finding lambda to right since E(f)=13.6 eV for the lyman series

Does this problem involve the Bohr model of the atom? In that case En = -13.6ev*Z^2/n^2

the energy at n = 1, comes out to: -13.6*5^2/1^2 = -340eV

I think your calculation was wrong: it should be -(1/2)*k(5e)e/r (this is the potential energy + kinetic energy)

and r =(5.29e-11 m)*(1)^2/(5)= 1.06e-11 m

using these the energy also comes out to: -340eV

I think the best way to get energy here is using En = -13.6ev*Z^2/n^2

So yes, get the energy difference from n=2 to n =1... then set that equal to hc/lambda... then solve for lambda.
 
learningphysics said:
Does this problem involve the Bohr model of the atom? In that case En = -13.6ev*Z^2/n^2

the energy at n = 1, comes out to: -13.6*5^2/1^2 = -340eV

I think your calculation was wrong: it should be -(1/2)*k(5e)e/r (this is the potential energy + kinetic energy)

and r =(5.29e-11 m)*(1)^2/(5)= 1.06e-11 m

using these the energy also comes out to: -340eV

I think the best way to get energy here is using En = -13.6ev*Z^2/n^2

So yes, get the energy difference from n=2 to n =1... then set that equal to hc/lambda... then solve for lambda.

Yes but doesn't E(0)=-13.6 eV only when we are talking about the hydrogen atom
 
  • #10
Benzoate said:
Yes but doesn't E(0)=-13.6 eV only when we are talking about the hydrogen atom

Ah... z = 6 not 5. sorry about that!

I think the bohr model is for 1-electron atoms... I understand now that carbon 5+ refers to a carbon ion... a carbon atom that has lost 5 of its electrons, and just left with 1...

like you found out carbon has atomic number 6. it lost 5 electrons, so it has just one left... So the bohr model still applies:

En = -13.6ev*Z^2/n^2

here if z = 1, then we're dealing with hydrogen... if we use Z = 6 then we have a carbon nucleus with 1 electron...

E_1 = -13.6*6^2/1^2 = -489.6eV

and using:

-(1/2)*k(6e)e/r will also give -489.6eV

so you can find E_2 etc...
 
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  • #11
learningphysics said:
Ah... z = 6 not 5. sorry about that!

I think the bohr model is for 1-electron atoms... I understand now that carbon 5+ refers to a carbon ion... a carbon atom that has lost 5 of its electrons, and just left with 1...

like you found out carbon has atomic number 6. it lost 5 electrons, so it has just one left... So the bohr model still applies:

En = -13.6ev*Z^2/n^2

here if z = 1, then we're dealing with hydrogen... if we use Z = 6 then we have a carbon nucleus with 1 electron...

E_1 = -13.6*6^2/1^2 = -489.6eV

and using:

-(1/2)*k(6e)e/r will also give -489.6eV

so you can find E_2 etc...

I still have two unknowns: E_2 and lambda. How would I find E_2? Why isn't E_1 -13.6 eV since the wavelength supposed to be in the lyman series

I think I have it : E_1=E_0/n^2= 13.6 eV/1^2=13.6 eV. E_2=E_0/n^2=13.6/2^2=3.4 eV. Only problems I see with this line of reason is how do you know the other transition state is n=2?
 
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  • #12
Benzoate said:
I still have two unknowns: E_2 and lambda. How would I find E_2? Why isn't E_1 -13.6 eV since the wavelength supposed to be in the lyman series

I think I have it : E_1=E_0/n^2= 13.6 eV/1^2=13.6 eV. E_2=E_0/n^2=13.6/2^2=3.4 eV.

this is only for hydrogen... for other atoms:

En = -13.6ev*Z^2/n^2

so for z = 6, E1 = -489.6 eV... using the same formula E2 = -122.4eV

Only problems I see with this line of reason is how do you know the other transition state is n=2?

I think that is just what "Lyman alpha transition" means.
 

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