GREPhysics.NET: GR8677 #26

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GR8677 #26

Problem

GREPhysics.NET Official Solution

Alternate Solutions

Verbatim question for GR8677 #26

Atomic $\Rightarrow$ }X-Rays

The K-series refers to the inner-most shell. (The order from inner to outer goes like K, L, M, N.) K-series refers to a transition from some outer state to the inner-most shell, where $n=1$ in the usual Bohr equation,

$E=-13.6(Z-1)^2 \left(1-1/n_i^2\right) eV, <br />$

where $(Z-1)^2$ is used to account for shielding.

For electrons bombarding a target, one assumes that the electrons are coming from $n_i=\infty$ , thus the equation becomes $E=-13.6(Z-1)^2\approx 10*900\approx 10,000$ eV, as in choice (D).

See below for user comments and alternate solutions!

Alternate Solutions

barefoot0
2006-11-14 11:17:02

I dont know if this exactly correct but I did the following.

I know 2.5eV gives ~ 500nm (This is just an easy thing to remember and quite usefull).

X rays are about .1 nm and E is inversly proportional to Lambda so

2.5eV*500nm/.1nm ~ 10,000 eV

Someone let me know if and why this method is wrong.

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Alternate Solution - Unverified

Comments

a19grey2
2008-11-03 20:29:42

Uh, am I the only person who calculates the value using $27^2$ and $n_i = 2$ , $n_f = 1$ and gets only ~7,400.

doing this on a test , I'd still guess the right answer, but this seems to far off for this to be the right answer since I got 7435.8eV using a calculator and on the test the approximations would make me even more off.

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NEC

FortranMan
2008-10-19 11:03:03

Shouldn't it be

$\left( \frac{1}{n^{2}_{i}} - \frac{1}{n^{2}_{f}} \right)$

Where n_{i} is the initial state (either at n_{i}=2 or $\infty$ ) and n_{f} is the final state (n_{f}=1)? This would explain how the negative cancels out.

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NEC

casaubon
2008-10-11 16:41:30

EDX (Energy Dispersive X-ray Spectroscopy) is a technique to characterize materials based on the wavelength of x-rays emitted from the sample when excited with an electron beam. Beam energy typically needs to be on the order of 10-25 keV to produce a good signal. So, choice D.

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NEC

eshaghoulian
2007-09-15 14:54:25

Agreed. The solution above is confusing the state of the bombarding electrons with the initial state of the electron making the transition. For minimum energy necessary, the equation is $E = (13.6eV)*((Z-1)^2)((1/1^2)-(1/n_{i}^2))$ . Minimizing this requires setting $n_{i}$ (the initial state of the transitioning electron) equal to 2 (note that you can't set it equal to 1 since then you wouldn't have a transition). This minimizes the energy necessary for the transition, and you equate this to the KE of the incoming electron, which isn't making any transitions. So the physical process is: electron comes in from $\infty$ , scatters off of an atom by giving up some of its KE, and then the atom makes a transition.

Note the general formula: $E = (13.6eV)*((Z-b)^2)((1/n_{f}^2)-(1/n_{i}^2))$ . This is a slight modification of Bohr's Law, where $b$ and $n_{f}$ depend on the series you are considering (e.g. $n_{f}=1$ DEFINES K series and gives $b=1$ , $n_{f}=2$ DEFINES L series and gives $b=7.4$ , etc.). The constants $b$ should probably be memorized, but I don't predict them asking beyond K or L series (probably won't even ask L series).

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Typo Alert!

Furious
2007-08-30 18:30:19

I have to agree with kevglynn, the minimum energy comes at n_i =2 not at n_i goes to infinity. That's actually the MAXIMUM energy.

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NEC

barefoot0
2006-11-14 11:17:02

I dont know if this exactly correct but I did the following.

I know 2.5eV gives ~ 500nm (This is just an easy thing to remember and quite usefull).

X rays are about .1 nm and E is inversly proportional to Lambda so

2.5eV*500nm/.1nm ~ 10,000 eV

Someone let me know if and why this method is wrong.

sonnb
2007-05-29 12:49:53

The spectrum of X rays ranges from 0.1nm to 10nm so that if you had used 1.0nm instead, you would have gotten the wrong answer.

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Alternate Solution - Unverified

kevglynn
2006-10-31 11:14:47

I'm pretty sure that the MINIMUM required energy refers to when the electrons come in from n_i=2. In other words, doesn't n_i refer to the initial state of the electron, not the bombarding electrons? thanks

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Typo Alert!

Healeyx76
2006-10-19 18:44:49

Is there a way to figure out R in Ev if you haven't memorized -13.6eV ? It is not one of the given values at the start of this test.

Oh: I guess you could memorize/know R = (m_e*e^4)/(2*hbar^2) eh?

13.6 13.6 13.6 13.6 ..... have to remember that one.

herrphysik
2006-10-27 21:56:12

Definately memorize that number.

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NEC

Post A Comment!

You are replying to:

Agreed. The solution above is confusing the state of the bombarding electrons with the initial state of the electron making the transition. For minimum energy necessary, the equation is $E = (13.6eV)*((Z-1)^2)((1/1^2)-(1/n_{i}^2))$ . Minimizing this requires setting $n_{i}$ (the initial state of the transitioning electron) equal to 2 (note that you can't set it equal to 1 since then you wouldn't have a transition). This minimizes the energy necessary for the transition, and you equate this to the KE of the incoming electron, which isn't making any transitions. So the physical process is: electron comes in from $\infty$ , scatters off of an atom by giving up some of its KE, and then the atom makes a transition. Note the general formula: $E = (13.6eV)*((Z-b)^2)((1/n_{f}^2)-(1/n_{i}^2))$ . This is a slight modification of Bohr's Law, where $b$ and $n_{f}$ depend on the series you are considering (e.g. $n_{f}=1$ DEFINES K series and gives $b=1$ , $n_{f}=2$ DEFINES L series and gives $b=7.4$ , etc.). The constants $b$ should probably be memorized, but I don't predict them asking beyond K or L series (probably won't even ask L series).

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