A beam of 5.0 MeV alpha particles (q=2 e) has a...

A beam of $5.0 \mathrm{MeV}$ alpha particles $(q=2 e)$ has a cross-sectional area of $1.50 \mathrm{~cm}^{2} .$ It is incident on flesh ( $\rho=950 \mathrm{~kg} / \mathrm{m}^{3}$ ) and penetrates to a depth of $0.70 \mathrm{~mm} .$ ( $a$ ) What dose (in Gy) does the beam provide to the flesh in a time of $3.0 \mathrm{~s}$ ? (b) What effective dose does it provide? Assume the beam to carry a current of $2.50 \times 10^{-9} \mathrm{~A}$ and to have $Q=14$. Using the current, find the number of particles deposited in the flesh in $3.0 \mathrm{~s}$, keeping in mind that for each particle $q=2 e:$ Number in $3.0 \mathrm{~s}=\frac{I t}{q}=\frac{\left(2.50 \times 10^{-9} \mathrm{C} / \mathrm{s}\right)(3.0 \mathrm{~s})}{3.2 \times 10^{-19} \mathrm{C}}=2.34 \times 10^{10}$ particles Each $5.0-\mathrm{MeV}$ alpha particle deposits an energy of $\left(5.0 \times 10^{6} \mathrm{eV}\right)\left(1.60 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)=8.0 \times 10^{-13} \mathrm{~J} .$ In $3.0$ s a total energy of $2.34 \times 10^{10}$ particles) $\left(8.0 \times 10^{-13} \mathrm{~J} /\right.$ particle $)$ is deposited. And it is delivered to a volume of area $1.50 \mathrm{~cm}^{2}$ and thickness $0.70 \mathrm{~mm}$. Therefore, $$ \text { Dose }=\frac{\text { Energy }}{\text { Mass }}=\frac{\left(2.34 \times 10^{10}\right)\left(8.0 \times 10^{-13} \mathrm{~J}\right)}{\left(950 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(0.070 \times 1.5 \times 10^{-6} \mathrm{~m}^{3}\right)}=188 \mathrm{~Gy}=1.9 \times 10^{2} \mathrm{~Gy} $$ Effective dose $=Q($ dose $)=(14)(188)=2.6 \times 10^{4} \mathrm{~Sv}$

A beam of $5.0 \mathrm{MeV}$ alpha particles $(q=2 e)$ has a cross-sectional area of $1.50 \mathrm{~cm}^{2} .$ It is incident on flesh ( $\rho=950 \mathrm{~kg} / \mathrm{m}^{3}$ ) and penetrates to a depth of $0.70 \mathrm{~mm} .$ ( $a$ ) What dose (in Gy) does the beam provide to the flesh in a time of $3.0 \mathrm{~s}$ ? (b) What effective dose does it provide? Assume the beam to carry a current of $2.50 \times 10^{-9} \mathrm{~A}$ and to have $Q=14$.

Using the current, find the number of particles deposited in the flesh in $3.0 \mathrm{~s}$, keeping in mind that for each particle $q=2 e:$
Number in $3.0 \mathrm{~s}=\frac{I t}{q}=\frac{\left(2.50 \times 10^{-9} \mathrm{C} / \mathrm{s}\right)(3.0 \mathrm{~s})}{3.2 \times 10^{-19} \mathrm{C}}=2.34 \times 10^{10}$ particles
Each $5.0-\mathrm{MeV}$ alpha particle deposits an energy of $\left(5.0 \times 10^{6} \mathrm{eV}\right)\left(1.60 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)=8.0 \times 10^{-13} \mathrm{~J} .$ In
$3.0$ s a total energy of $2.34 \times 10^{10}$ particles) $\left(8.0 \times 10^{-13} \mathrm{~J} /\right.$ particle $)$ is deposited. And it is delivered to a volume of area $1.50 \mathrm{~cm}^{2}$ and thickness $0.70 \mathrm{~mm}$. Therefore,
$$
\text { Dose }=\frac{\text { Energy }}{\text { Mass }}=\frac{\left(2.34 \times 10^{10}\right)\left(8.0 \times 10^{-13} \mathrm{~J}\right)}{\left(950 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(0.070 \times 1.5 \times 10^{-6} \mathrm{~m}^{3}\right)}=188 \mathrm{~Gy}=1.9 \times 10^{2} \mathrm{~Gy}
$$
Effective dose $=Q($ dose $)=(14)(188)=2.6 \times 10^{4} \mathrm{~Sv}$

Key Concepts

Effective Dose and Radiation Quality Factor

The effective dose takes into account the varying biological impact of different types of radiation by applying a quality factor (or radiation weighting factor) to the absorbed dose. This concept is used in radiation protection to provide a measure in Sieverts (Sv) that reflects not only the amount of energy absorbed, but also the potential for biological harm, thereby allowing a more comprehensive assessment of radiation risk.

Absorbed Dose Calculation (Gray)

The absorbed dose is defined as the amount of energy deposited per unit mass of a material, measured in Gray (Gy), where one Gray equates to one Joule per kilogram. This concept integrates the energy deposition information from the particles and the mass of the material in which the energy is deposited, serving as a quantitative measure of radiation exposure.

Relationship between Current, Charge, and Particle Number

This concept involves understanding that the electric current is the rate of charge flow, and when dealing with charged particles, one can calculate the number of particles passing a point by dividing the total charge delivered over a given time by the charge per particle. This is a basic principle in electrical and particle beam physics and is crucial for determining particle flux in experiments.

Energy Conversion from Electron Volts to Joules

Since particle energies are often measured in electron volts (eV) and standard energy calculations require Joules (J), this concept covers the conversion factor between these units. It is essential in physics to accurately determine the energy deposited by particles when translating individual particle energies into standard energy units suitable for further calculations.

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