Cutting the Dose Rate in half with shielding

[syllll_213]

Homework Statement

Which would halve Sarah's dose rate? Not Doubling the thickness of Perspex.

Relevant Equations

I = I0e(-x/lamdax)

Hi, I wonder how C should be calculated. I tried finding the ratio of new intensity to the original intensity, which gives the exponential chunck, and I wonder if that is enough to show that doubling the shield thickness would not halve the dose? Is there a clearer and more numerical representation?

 

[haruspex]

Sorry, but I cannot decipher your work. I (uppercase i)? 1? l (lowercase L)? Io? IO? I0? I₀?
All too confusing.

Suppose she is using a thickness $6.7N mm$ for some N, not necessarily an integer. What will her dose rate be? What will it be if she doubles the thickness? Is that / can that be half her present dose?

 

[Steve4Physics]

Homework Statement: Which would halve Sarah's dose rate? Not Doubling the thickness of Perspex.
Relevant Equations: I = I0e(-x/lamdax)

View attachment 362181View attachment 362180
Hi, I wonder how C should be calculated. I tried finding the ratio of new intensity to the original intensity, which gives the exponential chunck, and I wonder if that is enough to show that doubling the shield thickness would not halve the dose? Is there a clearer and more numerical representation?

For monochromatic gamma radiation and X-rays (but not $\alpha$ or $\beta$ radiation):
$I(d) = I_0 e^{-\mu d}$
where $\mu$ is the linear absorption coefficient and $d$ is the absorber thickness.

If $d$ is the ‘half-thickness’ then $I(d) = \frac 12 I_0$. That means $e^{-\mu d} = \frac 12$

If we have two half-thicknesses then:
$I(2d) = I_0 e^{-\mu 2d} = I_0( e^{-\mu d})^2 = I_0 \times (\frac 12)^2 = \frac 14 I_0$

BEGIN EDIT
The above shows that if the original shield thickness is the half-thickness, then doubling it does indeed halve the dose-rate (reducing it from $\frac 12 I_0$ to $\frac 14 I_0$).

But in this question we are give no information about the value of $d$. It may not be the half-thickness. If the original thickness, $d$, is some arbitrary value that reduces the intensity by a factor $f$ then a thickness of $2d$ reduces the intensity by a factor $f^2$ (e.g. from $0.60 I_0$ to $0.36 I_0$ which is not halving).
END EDIT

The exponential relationship is applicable to gamma/X-rays. It is not applicable to $\alpha$ or $\beta$ radiation (because $\alpha$ and $\beta$ patticles have finite ranges in matter). P-32 is a $\beta$ emitter so the above equations can’t be used. This may be an unintended error by the author of the question.

So (IMO) there is only one correct answer on the answer-list (not C).

Minor edit.

 

[jbriggs444]

It is not applicable to $\alpha$ or $\beta$ radiation (because $\alpha$ and $\beta$ patticles have finite ranges in matter). P-32 is a $\beta$ emitter so the above equations can’t be used.

If I am understanding correctly, this would be because the deceleration rate for an ionized particle is roughly proportional to velocity relative to the medium. So if it requires depth $d$ to reduce velocity by half, it will take additional depth $\frac{d}{2}$ to reduce velocity by half again at half the deceleration rate. One ends up summing a convergent geometric series to get total penetration to a depth of $2d$.

Very nice little fact.

 

[collinsmark]

[...]

If $d$ is the ‘half-thickness’ then $I(d) = \frac 12 I_0$. That means $e^{-\mu d} = \frac 12$

If we have two half-thicknesses then:
$I(2d) = I_0 e^{-\mu 2d} = I_0( e^{-\mu d})^2 = I_0 \times (\frac 12)^2 = \frac 14 I_0$

But here, I_0 would be the intensity with no shielding at all, correct?

But remember, she started out using some thickness of Perspex shielding to begin with. I think the question is asking about halving her dose rate compared her original thickness of shielding (not compared to no shielding at all).

The exponential relationship is applicable to gamma/X-rays. It is not applicable to $\alpha$ or $\beta$ radiation (because $\alpha$ and $\beta$ patticles have finite ranges in matter). P-32 is a $\beta$ emitter so the above equations can’t be used. This may be an unintended error by the author of the question.

Yes, this nugget of information makes the problem statement a bit suspect. (Or maybe it's a trick question and the student is supposed to know this about ^{32} \mathrm{P}.)

 

[Steve4Physics]

But here, I_0 would be the intensity with no shielding at all, correct?

Yes.

But remember, she started out using some thickness of Perspex shielding to begin with. I think the question is asking about halving her dose rate compared her original thickness of shielding (not compared to no shielding at all).

Good point. There is no indication of what the original thickness of shielding is. I (unjustifiably) assumed the original thickness is one half-thickness. I have edited my post.

Yes, this nugget of information makes the problem statement a bit suspect. (Or maybe it's a trick question and the student is supposed to know this about ^{32} \mathrm{P}.)

Not sure.