Warning

Lab 14 : Exercises, Questions & Answers

We have already practiced some math problems, see Lab 7: Ultrasound theory and there are some examples in the microscopy Lab 2 : Microscopy (optics), which was not part of the sylabus this year, but you might still find the exercises there useful. We will continue with some more problems today.

X-Rays

Useful formulas

• energy of a photon $E = hf = h \frac{c}{\lambda} \, ,$
• half-value layer $\frac{1}{2} = \textrm{e}^{-\mu d}$

Energy of a photon (I)

A photon with wavelength $100~nm$ has energy of $12~eV$, what is the energy of a photon with wavelength $2~nm$?

Energy of a photon (II)

The X-ray tube (rentgenka) generates X-rays from photons with kinetic energy of $10~keV$. Compute the wavelength of the generated X-rays if we know, that only 1% of the energy is converted to radiation.

Solution Compute the amount of the converted energy $E$, and then use it to get the wavelength.

X-rays with intensity of $10~W/cm^2$ passes through a $10~cm$ segment of tissue with half-value layer of $2~cm$. What will be the intensity after the tissue passage? What is the tissue density in Hounsfield units (HU), consider the linear attenuation of water to be $\mu_w = 0.22~cm^{-1}$? What kind of tissue is it?

Solution

We need the linear attenuation of the tissue $\mu_t$ to compute the density. We have $1/2 = \exp(-\mu_t \cdot d)$, which gives us $$\mu = \frac{\ln 2}{d} = \frac{\ln 2}{0.02} = 34.7 ~ \textrm{m}^{-1} = 0.347 ~ \text{cm}^{-1}$$

The intensity is then

$$I = I_0 \textrm{e}^{-\mu l} \approx 0.031 \cdot I_0$$

The Hounsfield units are defined as relative intensity w.r.t attenuation in water, in formula:

$$HU(t) = \frac{\mu_t - \mu_w}{\mu_w} \cdot k = \frac{0.347 - 0.22}{0.22} \cdot 10^3 \approx 575$$

So it is most likely a bone structure, as the result falls into the range [500, 1000] HU.

Alternative way We see, that the tissue segment is 5-times the half-width, which means we lose a half of the incident radiation on each 2 cm, so the final intensity is $$I = \frac{1}{2^5} \cdot I_0 = 0.03125 \cdot I_0$$.

Consider a tissue block, that contains $30~cm$ width of tissue A followed by a block of $8~cm$ of tissue B. Let the half-value layers be A: $10~cm$, B: $3~cm$. What is the intensity on the tissue boundary A|B? And what is the residual intensity of the exiting radiation?

Solution Again two possible solution ways. Either use the half-width formula to get the linear attenuation coefficients $\mu_A, \mu_B$ and then compute

$$I_{A|B} = I_0 \cdot \mathrm{e}^{-\mu_A l_A}$$

and

$$I_B = I_A \cdot \mathrm{e}^{-\mu_B l_B} = I_0 \cdot \mathrm{e}^{-\mu_A l_A - \mu_B l_B}$$

Or realize, that we pass 3-times the half-width when passing tissue A, so $$I_A = I_0 * (\frac{1}{2})^3$$ and $$I_B = I_A \cdot (\frac{1}{2})^{\frac{3}{8}}$$ to get the same result ($0.0197 \cdot I_0$)

Ultrasound

Doppler effect

Let us consider a doppler-US settings with carrier frequency of $3~MHz$ which we use to measure the speed of blood. Let the speed of blood at the imaging place be $2~cm/s$. What is the range of echoed frequencies we recieve from this spot? Use $c_{us} = 1540 m/s$.

Solution The situation can be separated in two parts – moving target (tissue) and static source (the US probe) and with the echo a moving source(tissue) and static target (the US probe), so the Doppler effect is twofold:

$$f = \frac{c \pm v}{c} \cdot \frac{c}{c \pm v}$$

Since the direction of the blood flow relative to the probe is unknown, we compute the minimal and maximal frequency

$$f_{min} = \frac{c-v}{c+v} \cdot f_0 \qquad f_{max} = \frac{c+v}{c-v} \cdot f_0$$

Their difference is $$f_{\Delta} = f_{max} - f_{min} = ... = \frac{4cv}{c^2 - v^2} \cdot f_{0} = 155.9 Hz$$

So we receive signals within $3~MHz \pm 78 Hz$.

Other

Useful equations

• Exponential decay $dN = -\lambda N dt$, where $N(0) = N_0$,
• the solution to whichis $N = N_0 \cdot \mathrm{e}^{-\lambda \cdot t}$, where $\lambda = \frac{\ln k}{t_k}$
• Decay constant and half-time $\lambda = \frac{\ln 2}{T_{1/2}}$
• Activity $A = \frac{dN}{dt}$

By which factor does the mass of a radioactive isotope reduce in 3 years, if it reduces four times within a year?

Solution Really need to see it?

Ok, we can look at the decay formula. At time $t$, we observe a certain amount of particles $N(t)$ (with initial amount N(0) = N_0) and know, that $N(t)$ changes – reduces – over time by the decay rate $\lambda$. We can express this relationship by the ordinary differential equation

$$dN = -\lambda N dt, \quad N(0) = N_0$$

The solution is

$$N = N_0 \cdot \mathrm{e}^{-\lambda \cdot t}, \qquad \lambda = \frac{\ln k}{t_k}$$

with decay ratio $k$ ($k=4$ in our case) and the duration $t_k$ ($t_k=1$ for us). This is a general formula, we typically use the half-time (time to reduce by factor 2), which is:

$$\lambda = \frac{\ln 2}{T_{1/2}}$$

Or just think and see, that we have a reduction by factor $(1/4)^3$ over the three years, so the mass reduces 64-times.

The initial decay rate (the activity) of $1~g$ mass of isotope $_{88}^{226}Ra$ is $1~Ci \approx 3.7 \cdot 10^{10} Bq$. What is the half-life? The molar mass of this isotope is $226 \cdot 10^{-3} kg \cdot mol^{-1}$.

Solution

As mentioned above, we derive the attenuation coefficient usually from the half-time $\lambda = \ln 2 / T_{1/2}$ which gives us $$N = N_0 \cdot \mathrm{e}^{-\lambda \cdot t}$$

Here, we look at the decay rate (activity $A$), which is the change of particles in time, formally

$$A = \left| \frac{dN}{dt} \right| = \lambda N(t) \quad = \frac{\ln 2}{T_{1/2}} \cdot N_0 \cdot \mathrm{e}^{- \frac{\ln 2}{T_{1/2}} \cdot t}$$

The exponential term equals 1 at time $t=0$, the initial amount of particles is given by the mass $m~[kg]$, the Avogadro constant $N_A = 6.023 \cdot 10^{23} [mol^{-1}]$ and the molar mass $M_m~[kg \cdot mol^{-1}]$

$$A(0) = \frac{\ln 2}{T_{1/2}} \cdot N_0 = \frac{\ln 2}{T_{1/2}} \cdot \frac{m \cdot N_A}{M_m}$$

We then reformulate the equation to obtain $T_{1/2} \approx 1583~[years]$.

Let us consider the usual PET radio-pharmaceutical with activity half-life of $130~[min]$ and half-life of elimination from the patient's body of $35~[min]$. The amount of $4\cdot 10^{-12}~[mol]$ of this pharmaceutical is produced $30~[min]$ before injection. What is the activity of at injection time? What is the activity after acquisition, which ends $15~[min]$ after injection?

Solution

The initial amount $N_0$ is given by the molar mass $n$ and $N_A$ – $N_0 = n \cdot N_A = 2.41 \cdot 10^{12} []$. Now we use $\lambda = \frac{\ln 2}{T_{1/2}}$ to get both elimination and activity decay constants $\lambda_E, \lambda_A$. It is

$$\lambda_E = \frac{\ln 2}{\tau_E} = \frac{\ln 2}{35 \cdot 60} = 3.3 \cdot 10^{-4} [s^{-1}]$$

and analogously $\lambda_A = 8.89 \cdot 10^{-5} [s^{-1}]$.

The injected amount $N_i$ is given by the decay rate formula as

$$N_i = N_0 \cdot \mathrm{e}^{-\lambda_A t_i} = ... = 2.05 \cdot 10^{12} []$$

To compute the activity, we need to compute the amount of particles available at time $t$ after injection (don't forget: the amount of particles is reduced by both elimination and activity), formally:

$$A(t) = \lambda_A \cdot N_i \mathrm{e}^{-(\lambda_E + \lambda_A) t}$$

The initial activity is then $A(0)$, and the activity after acquisition $A(15~[min])$ with results $183~[MBq]$ and $125~[MBq]$ respectively.

PET

Let us have a PET scanner with detector ring with diameter of $1~[m]$ formed by $N = 200$ detectors with equal size. Further, let us have the source of radioactivity placed at the exact center of the detector ring with the total activity of $A = 10^6~[Bq]$ (the activity registered by all detectors). In case a decay event occurs, what is the probability of event detection by a specific pair of detectors? What is the delay time between decay and detection? What is the expected activity at time $T=10~[min]$ if the radioactive agent has a half-life of $\tau_A = 10~[min]$ and the elimination half-life is also $\tau_E = 10~[min]$? What is the number of detected events from beginning until $T = 10~[min]$? Compute the total number of events and the number of events per detector pair.

Solution

Since the agent is placed exactly at the center of the ring, the detection lines will pass through the center. That means, we have $N/2$ possible detector pairs and the detection probability is $\frac{1}{100}$.

The gamma ray spreads with speed of light, so the delay between event and detection is $$t = \frac{0.5 * d}{c} = \frac{0.5}{3\cdot 10^8} = 1.67 \cdot 10^{-9}~[s]$$

If we look precisely at the input values we see, that we need to estimate the activity exactly the half time value. So we lose half of the particles due to activity and from them another half that gets eliminated from the body. So in total a $\frac{1}{4}$ of the original activity will remain. Thus $A(T=10~[min]) = 2.5 \cdot 10^5~[Bq]$.

So the amount of active particles changed from $N_0$ to $\frac{1}{4} \cdot N_0$ during the scan time of $10~[min]$. Since half of them was eliminated from the body, we arrive at $$\delta N = \frac{1}{2} \cdot (N_0 - \frac{1}{4}\cdot N_0) = \frac{3}{8} N_0$$ decay events (total count).

If we want to get the number, we need to compute the decay constant $\lambda = \frac{\ln 2}{\tau_{1/2}} = 1.16 \cdot 10^{-3}$ and put it in $N_0 = \frac{A_0}{\lambda}$. This will give $N_0 = 862 \cdot 10^6~[.]$