# Calculation of flux/

1. Nov 6, 2009

### Rajini

hello all,
i wanted to calculate the illumination at counter using inverse square law..(not a home work problem)
In my case i have a flat radioactive source (not point source) with a surface area of $$\pi(2.5)^2$$mm$$^2$$. Source ($$^{57}$$Co) strength is $$1.85\times 10^9$$Bq. Half-life=272 days. Now what will be the flux arriving at the detector placed 300 mm far from the source?
Second question: To convert Ci to Bq, we just multiply by $$3.7\times10^{10}$$. Is there any other conversion to convert Ci to some 'X' by dividing by 9 ??
If some one knows some links/url/notes for calculating the count rate please reply me..
I dont how to apply the inverse $$^2$$ law!!!
Thanks for your reply.

2. Nov 6, 2009

### Bob S

You need to specify what flux you want.

1 Ci (Curie) is 3.7 x 1010 decays per second into 4 pi sterads.You need to multiply this value by the number and branching ratio of all the photons (and charged particles) to get count rate.

Bob S

[Added] Here is what the following URL says about Co-57 branching ratios:
Appendix I. The Cobalt-57 radioactive decay.

Cobalt-57 decays to iron-57 by beta decay, with a half-life of about 270 days. The energy-level scheme for iron-57 is shown in Figure AI-1 below. The decay almost always (99.8% of the time) goes to the JP = 5/2- excited state of the iron-57 nucleus. And most of the time this state decays in two steps, giving gamma rays of 122 keV and 14 keV. The 14-keV gamma ray is the line used to observe Mossbauer absorption. In 11% of decays the state goes directly to the ground state, giving a 136 keV gamma ray.

In addition to these three gamma rays, the iron-57 atom will emit abundant K*a and Kb atomic X-rays, at about 6 keV. Thus the dominant features of the spectrum from a sodium-iodide detector should be clear peaks at 6, 14 and 122 keV, with a weak line at 136 keV.

See http://www.physics.sfsu.edu/~bland/courses/490/labs/Mossbauer/mossbauer.html [Broken]

It is an EC decay
the gamma branching ratios are about
14.4 KeV 89%
122 kev 89%
136 kev 11%

Bob S

Last edited by a moderator: May 4, 2017
3. Nov 7, 2009

### Rajini

Hi Bob,
thanks for you reply. From you reply i got one important message that accounts for '9'.
Okay, i actually work with MÃ¶ssbauer spectroscopy (only the 14.4 keV level, which is of 9.2% of the total decay of 57Co). So for making calculation for the flux that arrive at detector one need to divide source strength by that 9.2. Am i correct?????.
What is your opinion??

Ps:The values you gave is actually the sum of conversion electron and gamma decay probability. Also i don't know why in that website they wrote that 57Co decay process is beta decay...it is 100% EC.
thanks.

4. Nov 7, 2009

### Bob S

Here is the best information I can find on the Cobalt-57 branching ratios (see thumbnail). The 14.4 KeV gamma is about 9.15%. This state seems to have a high probability of decaying by auger(?) electrons plus K x-rays.
Bob S

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5. Nov 11, 2009

### bcrowell

Staff Emeritus
The 2.5 mm radius of your source is very small compared to the 300 mm distance from source to detector, so I think you'd get an excellent approximation without worrying about the complications from the non-pointlike source geometry. What kind of precision do you need? At the level of precision where you were worrying about the non-pointlike source, you'd probably also need to worry about the finite size and detailed geometry of the detector.

6. Nov 11, 2009

### Rajini

Hi, Yes what you said is correct..
But the source strength is divided by 9 or 9.2? Instead of multiplying (as bob said: it should be multiplied).
PS: detector counts only the events of 14.4 keV, which is roughly 9 or 9.2 per 100 disintegration. Source strength is 1.85X10^9 disintegration per second.
thanks

Last edited: Nov 11, 2009
7. Nov 11, 2009

### Bob S

I think the conversion factor is 3.7 x 1010 as you show. Both Ci and Bq relate to disintegrations per second into 4 pi sterads. So neither is disintegrations per sterad. Secondly, it is all disintegrations, and not just gammas or betas.
Bob S

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