# Homework Help: New distance between the Geiger counter and the source

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1. Oct 28, 2016

### moenste

1. The problem statement, all variables and given/known data
A point source of γ-radiation has a half-life of 30 minutes. The initial count rate, recorded by a Geiger counter placed 2.0 m from the source, is 360 s-1. The distance between the counter and the source is altered. After 1.5 hour the count rate recorded is 5 s-1. What is the new distance between the counter and the source?

2. The attempt at a solution
We have:
T1 / 2 = 1800 s.
d = 2 m
R = 360 s-1

t = 5400 s
d = ?
R = 5 s-1

No idea what to do here. R = R0 e- λ t, but I don't see a way to find distance from this formula. There is also a formula from the experimental verification of the inverse square law for γ-rays: x = k R- 1 / 2 - c, where d = x + c (distance between actual position of source and effective point of detection), k = constant of proportionality. But I still see not much connection between these two formulas.

Maybe I'm looking in the wrong direction?

2. Oct 28, 2016

### Staff: Mentor

The inverse square law applies. First determine the new count rate that would be detected at the original distance (if the counter were not moved). Any difference of the actual measured rate from that rate must be due to the change in distance from the source.

3. Oct 28, 2016

### Hamal_Arietis

We seem number of photons will be distributed onto a sphere, If N is number of photons that appears at time t, $\frac{N}{4\pi r^2}$ will be the number of photons that the count rate recorded.

4. Oct 28, 2016

### moenste

R = R0 e- λ t = 360 * e- (ln 2 / 1800) * 5400 = 45 s-1?

It looks like the difference is 45 - 5 = 40 s-1.

I also simplified the formula d = k / √ R. Though I've no idea what is k.

5. Oct 28, 2016

### Staff: Mentor

Good.
Differences won't help. Use ratios to scale the rate with distance. It's an inverse square law so rate ∝ 1/d2.

6. Oct 28, 2016

### moenste

I ∝ 1 / d2
I ∝ 1 / 22
I ∝ 0.25

I still don't the use of it. Maybe you mean 45 s-1 / 2 m = 5 s-1 / d → d = 0.22 m?

7. Oct 28, 2016

### Hamal_Arietis

At intial time, the count rate recorded is R. It means:
$$R=\frac{N_0}{4\pi d^2}$$
The distane d will change into d'. After 1,5h and $N_0$ will change into N. Find the ratio d'/d. You will have the answer

8. Oct 28, 2016

### Staff: Mentor

Suppose you have a quantity A that varies as the inverse square of the distance d. Then inserting a constant of proportionality k you can write:

$A = \frac{k}{d^2}$

Now if you have two pairs of data, $(A_1,d_1)$ and $(A_2,d_2)$, then you can write:

$A_1 = \frac{k}{d_1^2}$ and $A_2 = \frac{k}{d_2^2}$

You can then form the ratios:

$\frac{A_2}{A_1} = \frac{\frac{k}{d_2^2}}{\frac{k}{d_2^2}} = \frac{d_1^2}{d_2^2}$

Or, extracting the important relationship:

$\frac{A_2}{A_1} = \frac{d_1^2}{d_2^2}$

If you have a known data pair and one half of another data pair, you can find the "missing" value.

9. Oct 28, 2016

### Hamal_Arietis

Sorry my equation is incorrect.
R is the number of photon that the count rate recorded in 1s. So at the time A point source of γ-radiation has $N_0$ photons. Number of photons that count rate recorded in 1s is:
$$R=\frac{N_0(1-e^{-1.\lambda })}{4\pi d^2}$$
After 1,5h and d change into d'
$$R'=\frac{N(1-e^{-1.\lambda })}{4\pi d'^2}=\frac{N_0.e^{-\lambda t}.(1-e^{-1.\lambda })}{4\pi d'^2}$$

Find the ratio.

10. Oct 28, 2016

### moenste

Get it, just rearrange the equation for k and then substitute in the first equation parts of the second instead of k.

R1 d12 = R2 d22
45 * 22 = 5 * d22
d2 = 6 m.

Thank you!