We have already practiced some math problems, see Lab 05: Ultrasound theory and there are some examples in the X-rays Lab 09: X-ray. We will continue with some more problems today.

Important formulas

• energy of a photon $E = hf = h \frac{c}{\lambda} \, ,$
• half-value layer $\frac{1}{2} = \textrm{e}^{-\mu d}$
• speed of light in vacuum $c \approx 3 \cdot 10^8 m \cdot s^{-1}$
• Planck's constant $h \approx 4.135 \cdot 10^{-15} eV \cdot Hz^{-1}$

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

Solution Let $E_1 = 12 eV$ the energy of the photon with wavelength $\lambda_1 = 100 nm$ and the question is what is the energy $E_2$ of a photon with wavelength $lambda_2 = 2 nm$.

We can formulate the relationship as $E_1 \cdot \lambda_1 = h\cdot c = E_2 \cdot \lambda_2$ from $E = \frac {h \cdot c} {\lambda}$

Then $E_2 = E_1 \frac{\lambda_1}{\lambda_2} = 12 \frac{100 \cdot 10^{-9}}{2 \cdot 10^{-9}} = 600 eV$

The X-ray tube (rentgenka) generates X-rays from electrons 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 $E' = 0.01 \cdot E = 100eV$, and then use it to get the wavelength from $\lambda = \frac {h \cdot c} {E'} = 12.4 \cdot 10^{-9} m$.

X-rays with intensity of $10~W/cm^2$ passes through a $10~cm$ segment of tissue with a half-value layer of $2~cm$. What will be the intensity after the tissue passage? What is the tissue density in Hounsfield units (HU), considering 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:

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

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 ways to the solution. 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 ($0.0197 \cdot I_0$)

Important formulas

Doppler effect $ f = \frac {c \pm v_r} {c \pm v_s} \cdot f_0$, where $f_0$ is the emitted frequency, $f$ is the observed frequency, $c$ is the propagation speed of the waves, $v_r$ is the speed of the receiver and $v_s$ is the speed of the source; $v_r$ is added (+) when the source moves towards the receiver and subtracted (-) when the source moves away from the receiver, $v_s$ is added (+) when the receiver moves away from the receiver and subtracted (-) when the receiver moves towards the receiver.

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

Solution The situation can be separated into 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 $3~MHz \pm 78 Hz$.

Important equations

• Exponential decay $ dN = -\lambda N dt$, where $N(0) = N_0$,
• the solution to which is $ 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

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

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

So 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 a factor 2), which is:

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

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 $T_{1/2} \approx 1583~[years]$.

A sample of $_{18}F$ is measured at 10:40 and has an activity of 30 MBq.It is injected into a patient at 11:30. How much activity was injected? The half-life of $_{18}F$ is 109.8 min.

Solution

The activity at the time of injection is $ A(t) = 30 \cdot 10^6 \cdot \mathrm{e}^{- \frac{ln 2 \cdot 50} {109.8}} = 21.9 \cdot 10^6 Bq$

Let us consider the usual PET radiopharmaceutical with an activity half-life of $130~[min]$ and a half-time 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 the radiopharmaceutical 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.

Let us have a PET scanner with a detector ring with a diameter of $1~[m]$ formed by $N = 200$ detectors of equal size. Further, let the source of radioactivity be placed at the exact center of the detector ring, with a 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 the 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 the speed of light, so the delay between the 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 at the half-time value. So we lose half of the particles due to activity, and of those, another half is 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 were 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 $N_0 = 862 \cdot 10^6~[.]$ the total number of particles at the beginning, the number of detected events is $325 \cdot 10^6~[.]$, which is approximately $3.24 \cdot 10^6~[.]$ per detector pair.

Important equations

• konvenční zraková vzdálenost optical distance $d = 250 mm$
• zvětšení magnification $Z = \frac{y'}{y} = -\frac{a'}{a}$
• zobrazovací rovnice spojné čočky lense maker's formula $\frac{1}{a} + \frac{1}{a'} = \frac{1}{f}$ with $a, a'$ the object and image distance, $f$ is the focal length
• zvětšení mikroskopu magnification of a microscope

$$ Z = \frac{\Delta}{f_{ob}} \frac{d}{f_{oc}} $$

• Abbé-Rayleigh criterion $d = 1.22 \cdot \frac{\lambda}{2 \cdot NA}$

Let us consider a thin convex lens (spojná čočka) with a focal length of 10 cm. Find the magnification for a red object of height 6 cm, placed at a distance of 12 cm in front of the imaging plane.

Solution Given: $f=0.1~m, y=0.06~m, a=0.12~m$ Magnification is the ratio of the size of the displayed object to the size of the original object $Z = \frac{y'}{y} = \frac{a'}{a}$ and the imaging equation gives us $a' = \frac{f \cdot a}{a - f}$ and finally, after substituting $a'$, we get $$Z = \frac{f}{a - f} = \frac{0.1}{0.12 - 0.1} = 5$$

Consider a simple microscope formed by two thin convex lenses with an objective focal length of 5 mm and a diameter of 1 cm, a tube optical length of 100 mm, and an eyepiece focal distance of 25 mm. Draw a schematic picture of the microscope and compute its magnification.

Solution Given is $f_\text{ob} = 0.005$, $f_\text{oc} = 0.025$, tube length $\Delta = 0.1 m$ and conventional optical distance $d=0.25 $m. By simple insertion into the magnification equation, we get

$$ Z = \frac{\Delta}{f_{\text{ob}}} \cdot \frac{d}{f_{\text{oc}}} = \frac{0.1}{0.005} \frac{0.25}{0.025} = 200. $$

(a) What magnification is produced by a 0.150 cm focal length microscope objective that is 0.155 cm from the object being viewed?

(b) What is the overall magnification if an 8× eyepiece (one that produces a magnification of 8.00) is used?

Solution (a) From the lens maker's formula $\frac{1}{a} + \frac{1}{a'} = \frac{1}{f}$ we get $a' = \frac{af}{a-f}$ and when plugged into the formula for the magnification, we get $Z = \frac{f}{a-f} = \frac{0.15 cm}{0.155 cm - 0.15 cm} = 30$

(b) The total magnification is the product of the objective magnification and the eyepiece magnification, $Z_{tot} = Z_{obj} \cdot Z_{eye} = 30 \cdot 8 = 240$.

Consider a simple microscope with an objective magnification of 40, eyepiece magnification of 10, and tube length of 12 cm. Compute the magnification as well as the focal lengths of the objective and the eyepiece.

Solution

The magnification is:

$$ Z = \frac {\Delta} {f_{ob}} \cdot \frac {d} {f_{oc}} = Z_{ob} \cdot Z_{oc} = 40 \cdot 10 = 400 $$

The focal lengths:

$$ f_{ob} = \frac {\Delta} {Z_{ob}} = \frac {0.12} {40} = 0.003~m $$ $$ f_{oc} = \frac {d} {Z_{oc}} = \frac {0.25} {10} = 0.025~m$$

Let us consider a microscope with an eyepiece numerical aperture (NA) of 0.4, using illumination by light of wavelength 650 nm. Compute the maximal resolution (i.e., the minimal distance between distinguishable points) we can achieve in this setting.

Solution From Abbe-Rayleigh criterion: $$ d = 1.22 \frac{\lambda}{2 NA} = 1.22 \cdot \frac{6.5 \cdot 10^{-7}}{2 \cdot 0.4} = 990 ~n \text{m} = 1 ~\mu \text{m} \, $$

Let's consider the following microscope objectives (taken from Microscope objectives) and their numerical apertures NA.

In your task, it is crucial to observe 500 nm details using standard lightning ($\lambda$ = 650 nm) and no special lens immersion ($n_{air}$ = 1).

(a) Which one of the objectives would offer you the desired resolution?

(b) Select one of the applicable objectives and compute the half-cone angle $theta$.

Solution We can compute the resolution of the objectives using the Abbé-Rayleigh criterion: $d = 1.22 \frac{\lambda}{2\text{NA}}$. See the table below for the computed values. The objectives Lambda 40XC and Lambda 60XC have a resolution better than 500 nm. The half-cone angle can be devised from the formula $\text{NA} = n \cdot \text{sin} \theta \rightarrow \theta = \text{arcsin} \frac{\text{NA}}{n} = \text{arcsin} \frac{0.95}{1} = 1.25~\text{rad} = 71.8˚$.

Let us consider a lens $L$ with a half-cone angle $\theta=30^\circ$ in water ($n = 1.33$) and normal light illumination ($\lambda$ = 650 nm). Which light (wavelength and color) do we need to use to get the same resolution capabilities when using air ($n=1$) as a diffraction medium?

Solution

Numerical aperture $\text{NA} = n \sin \theta$. Same resolution means

\begin{align} d_1 &= d_2 \\ 1.22\cdot \frac{\lambda_1}{2 \cdot \text{NA}_1} &= 1.22 \cdot \frac{\lambda_2}{2 \cdot \text{NA}_2} \\ \frac{\lambda_1}{n_1 \sin \theta} &= \frac{\lambda_2}{n_2 \sin \theta} \\ \lambda_2 &= \lambda_1 \cdot \frac{1}{1.33} \\ \lambda_2 &= \frac{650}{1.33} = 488 \text{nm} \end{align}

Therefore, we need blue light illumination.