Module Questions 21 and 22


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Or Fisher 1I
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Joined: Fri Feb 02, 2018 3:00 am

Module Questions 21 and 22

Postby Or Fisher 1I » Tue May 08, 2018 9:30 pm

Please help. I keep doing these two questions from the modules and I get the same wrong answer every time. For 21 I get C and for 22 I get D but those are incorrect. Does anyone know how to solve these? Thanks.

21. The electron is not confined to the nucleus and we now know that the size of an atom is determined by its electrons outside of the nucleus. For hydrogen its measured atomic diameter is 145,000 times its nuclear diameter of 1.7 x 10-15 m. In other words make the (correct) assumption that the electron is confined to the atomic diameter of 2.5 x 10-10 m which would be a better estimate of the electron's uncertainty in position. Repeat all of the above calculations and comment on your values. Use the Heisenberg uncertainty equation to calculate the electron's uncertainty in momentum. Then use the mass of electron 9.1 x 10-31 kg to calculate its uncertainty in velocity. Comment on your value.

A. Delta p <= 25 kg.m.s-1, Delta v = 4.3 x 10-6 m.s-1, A small number but at least it is physically possible.

B. Delta p <= 25 kg.m.s-1, Delta v = 2.3 x 105 m.s-1, A large number but at least it is physically possible.

C. Delta p >= 2.1 x 10-25 kg.m.s-1, Delta v = 4.3 x 10-6 m.s-1, A small number but at least it is physically possible.

D. Delta p >= 2.1 x 10-25 kg.m.s-1, Delta v = 2.3 x 105 m.s-1, A large number but at least it is physically possible.

E. None of the above

22. Use the above uncertainty in velocity (correct model) to calculate the electron's uncertainty in kinetic energy. Then calculate the uncertainty in kinetic energy per mole of electrons (that is, per mole of hydrogen atoms). Comment on your value.

A. 2.4 x 10-20 J/e, 1.4 x 104 J/mol, These more reasonable numbers indicate that this is a more stable and realistic model.

B. 2.4 x 10-20 J/e, 5.1 x 10-18 J/mol, These less reasonable numbers indicate that this is a less stable and unrealistic model.

C. 8.4 x 10-42 J/e, 1.4 x 104 J/mol, These more reasonable numbers indicate that this is a more stable and realistic model.

D. 8.4 x 10-42 J/e, 5.1 x 10-18 J/mol, These less reasonable numbers indicate that this is a less stable and unrealistic model.

E. None of the above

Bryan Jiang 1F
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Joined: Fri Apr 06, 2018 11:03 am
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Re: Module Questions 21 and 22

Postby Bryan Jiang 1F » Wed May 09, 2018 12:33 am

I calculated the answers to be D and A respectively.

21. ΔpΔx ≥ h/4π. Heisenberg’s Indeterminancy Equation.

Δx = 2.5 x 10^-10 m. Given.

Δp(2.5 x 10^-10) ≥ h/4π. Plug Δx into Heisenberg’s Indeterminancy Equation.

Δp ≥ 2.1 x 10^-25 kg.m.s^-1. Answer as electron’s uncertainty in momentum.

———

Δp = mΔv. Uncertainty in momentum is mass multiplied by uncertainty in velocity.

= (9.1 x 10^-31 kg)(Δv). Mass of electron is given to plug in.

(9.1 x 10^-31 kg)(Δv) ≥ 2.1 x 10^-25 kg.m.s^-1. Plug mΔv into previous answer.

Δv ≥ 2.3 x 10^5 m.s^-1. Answer as electron’s uncertainty in velocity.

———

“A large number but at least it is physically possible.” An appropriate comment since an uncertainty of 2.3 x 10^5 m.s^-1 is indeed large as it is close to c = 3.8 x 10^8 m.s^-1, and it is physically possible since the velocity would still be less than c.

D.
———

22. ΔE kinetic = ½mΔv². Electron’s minimum uncertainty in kinetic energy.

ΔE kinetic = ½(9.1 x 10^-31 kg)(2.3 x 10^5 m.s^-1)². Plug in given value for electron’s mass and previously calculated value for uncertainty in electron’s velocity. Do NOT forget to square the velocity value!

ΔE kinetic = 2.4 x 10^-20 J. Uncertainty in kinetic energy per electron.

———

2.4 x 10^-20 * 6.022 x 10^23. Multiply uncertainty in kinetic energy per electron by Avogadro’s number to find the uncertainty in kinetic energy per mole of electrons, or per mole of hydrogen atoms.

= 1.4 x 10^4 J. Answer as the uncertainty in kinetic energy per mole of electrons, or per mole of hydrogen atoms.

———

“These more reasonable numbers indicate that this is a more stable and realistic model.” An appropriate comment since that amount of energy per mole is conceivable.

A.


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