===== Lab 3 : X-rays, radioactive decay ===== ==== 1 X-Rays ==== **Useful formulas** * energy of a photon $E = hf = h \frac{c}{\lambda}$ * linear attenuation $I = I_0 \cdot e^{-\mu d}$ * mass attenuation coefficient $\mu_m = \frac{\mu}{\rho_m}$ * 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 $h \approx 4.135 \cdot 10^{-15} eV \cdot Hz^{-1}$ === 1.1 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 $2~nm$? [600 eV] /* **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 $E_2 = E_1 \frac{\lambda_1}{\lambda_2} = 12 \frac{100 \cdot 10^{-9}}{2 \cdot 10^{-9}} = 600 eV$ */ === HW 1.2 Energy of a photon (II) [1 pt] === 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 $\lambda = \frac {h \cdot c} {E'} = 12.4 \cdot 10^{-9} m$. */ === 1.3 Radiation absorption (I) === 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? For typical Hounsfield units ranges of tissues you can refer to [[https://en.wikipedia.org/wiki/Hounsfield_scale#Values_for_different_body_tissues_and_material|wiki]]. [575, bone] /* **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 $$I = \frac{1}{2^5} \cdot I_0 = 0.03125 \cdot I_0$$. */ === HW 1.4 Radiation absorption (II) [1 pt] === 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 ($0.0197 \cdot I_0$) */ === HW 1.5 Linear attenuation coefficient[1pt] === The linear attenuation coefficient for an unknown material shall be determined. The following data are obtained in a measurement made in “narrow beam geometry”. Material thickness: 2.5 cm. Measurement without material: 35 000 counts during 300 s (including background). Measurement with material: 25 700 counts during 300 s (including background). Measurement of the background (both measurements): 2350 counts during 600 s. Hint: The count rate r follows similar relationships like the activity; $r_d = r_0 e^{-\mu d}$ where $\mu$ is the linear attenuation coefficient, $r_0$ is the net count rate before attenuation and $r_d$ is the net count rate after the attenuation. /* **Solution** The linear attenuation coefficient is given as $\mu = \frac{-ln(\frac{r_d}{r_0})}{d}$. The net count rate before attenuation is $r_0 = \frac{35 000}{300} - \frac{2 350}{600} = 112.75~s^{-1}$ and the net count rate after attenuation is $r_d = \frac{25 700}{300} - \frac{2 350}{600} = 81.75~s^{-1}$. Altogether, the linear attenuation coefficient is $\mu = \frac{-ln(\frac{112.75}{81.75})}{2.5} = -0.129~cm^{-1}$ */ === HW 1.6 Mass attenuation coefficient [1 pt] === What fraction of 140 keV x‐rays incident upon a 0.5 mm thick lead apron will be transmitted? The mass absorption coefficient of lead for 140 keV x‐rays is: $μ_m = 2.0~cm_2 \cdot g^{‐1}$ and the density of lead is $\rho = 11.3~g \cdot cm^{-3}$. /* **Solution** We want to know the linear attenuation, therefore we have to convert the mass attenuation coefficient to linear attenuation using the relationship $\mu_m = \frac{\mu}{\rho_m}$. The linear attenuation coefficient is $\mu = \mu_m \cdot \rho = 2.0~cm_2 \cdot g^{‐1} \cdot 11.3~g \cdot cm^{-3} = 22.6~cm^{-1}$. We get the fraction of the x-rays intensity by using the equation for linear attenuation $I = I_0 \cdot e^{-\mu d} = I_0 \cdot e^{-22.6~cm^{-1} \cdot 0.05~cm} = 0.323 \cdot I_0$. The fraction of x-rays intensity that is transmitted through the lead apron is 0.323. */ ==== 2 Radioactive decay ==== ** Useful 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}$ * Avogadro number $N_A = 6.022 \cdot 10^{23} mol^{-1}$ === 2.1 Radioactivity === By which factor does the mass of a radioactive isotope reduce in 3 years, if it reduces four times within a year? [$\frac{1}{64}=0.015625$] /* **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 $$ dN = -\lambda N dt, \quad N(0) = N_0 $$ 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: $$\lambda = \frac{\ln 2}{T_{1/2}}$$ */ === HW 2.2 Radioactive decay === 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 $T_{1/2} \approx 1583~[years]$. */ === 2.3 Radiopharmaceuticals I === 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. [$21.9 \cdot 10^6 Bq$] /* **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$ */ === BONUS 2.4 Radiopharmaceuticals II === Let us consider the usual PET radiopharmaceutical 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 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. */