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Schaum’s Outline of College Physics

Eugene Hecht

Chapter 46

Applied Nuclear Physics - all with Video Answers

Educators

Chapter Questions

02:40

Problem 1

The binding energy per nucleon for ${ }^{238} \mathrm{U}$ is about $7.6 \mathrm{MeV}$, while it is about $8.6 \mathrm{MeV}$ for nuclei of half that mass. If a ${ }^{238} \mathrm{U}$ nucleus were to split into two equal-size nuclei, about how much energy would be released in the process?

There are 238 nucleons involved. Each nucleon will release about $8.6-7.6=1.0 \mathrm{MeV}$ of energy when the nucleus undergoes fission. The total energy liberated is therefore about $238 \mathrm{MeV}$ or $2.4 \times 10^{2} \mathrm{MeV}$.

Km Neeraj
Km Neeraj
Numerade Educator
04:03

Problem 2

What is the binding energy per nucleon for the ${ }_{92}^{238} \mathrm{U}$ nucleus? The atomic mass of ${ }^{238} \mathrm{U}$ is $238.05079 \mathrm{u} ;$ also $m_{p}=1.007276 \mathrm{u}$ and $m_{n}=1.008665 \mathrm{u}$.
The mass of 92 free protons plus $238-92=146$ free neutrons is
$$
\text { (92) }(1.007276 \mathrm{u})+(146)(1.008665 \mathrm{u})=239.93448 \mathrm{u}
$$
The mass of the ${ }^{238} \mathrm{U}$ nucleus is
$$
238.05079-92 m_{e}=238.05079-(92)(0.000549)=238.00028 \mathrm{u}
$$
The mass lost in assembling the nucleus is then
$$
\Delta m=239.93448-238.00028=1.9342 \mathrm{u}
$$
Since $1.00 \mathrm{u}$ corresponds to $931 \mathrm{MeV}$,
$$
\text { Binding energy }=(1.9342 \mathrm{u})(931 \mathrm{MeV} / \mathrm{u})=1800 \mathrm{MeV}
$$
and
$$
\text { Binding energy per nucleon }=\frac{1800 \mathrm{MeV}}{238}=7.57 \mathrm{MeV}
$$

Km Neeraj
Km Neeraj
Numerade Educator
05:27

Problem 3

When an atom of ${ }^{235} \mathrm{U}$ undergoes fission in a reactor, about $200 \mathrm{MeV}$ of energy is liberated. Suppose that a reactor using uranium- 235 has an output of $700 \mathrm{MW}$ and is 20 percent efficient.
(a) How many uranium atoms does it consume in one day? ( $b$ ) What mass of uranium does it consume each day?
(a) Each fission yields
$$
200 \mathrm{MeV}=\left(200 \times 10^{6}\right)\left(1.6 \times 10^{-19}\right) \mathrm{J}
$$
of energy. Only 20 percent of this is utilized efficiently, and so
Usable energy per fission $=\left(200 \times 10^{6}\right)\left(1.6 \times 10^{-19}\right)(0.20)=6.4 \times 10^{-12} \mathrm{~J}$
Because the reactor's usable output is $700 \times 10^{6} \mathrm{~J} / \mathrm{s}$, the number of fissions required per second is
$$
\text { Fissions/s }=\frac{7 \times 10^{8} \mathrm{~J} / \mathrm{s}}{6.4 \times 10^{-12} \mathrm{~J}}=1.1 \times 10^{20} \mathrm{~s}^{-1}
$$
and Fissions/day $=(86400 \mathrm{~s} / \mathrm{d})\left(1.1 \times 10^{20} \mathrm{~s}^{-1}\right)=9.5 \times 10^{24} \mathrm{~d}^{-1}$
(b) There are $6.02 \times 10^{26}$ atoms in $235 \mathrm{~kg}$ of uranium- $235 .$ Therefore, the mass of uranium- 235 consumed in one day is
$$
\text { Mass }=\left(\frac{9.5 \times 10^{24}}{6.02 \times 10^{26}}\right)(235 \mathrm{~kg})=3.7 \mathrm{~kg}
$$

Abid Hussain
Abid Hussain
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03:32

Problem 4

Neutrons produced by fission reactions must be slowed by collisions with moderator nuclei before they are effective in causing further fissions. Suppose an $800-\mathrm{keV}$ neutron loses 40 percent of its energy on each collision. How many collisions are required to decrease its energy to $0.040 \mathrm{eV} ?$ (This is the average thermal energy of a gas particle at $35^{\circ} \mathrm{C}$.)

After one collision, the neutron energy is down to $(0.6)(800 \mathrm{keV})$. After two, it is $(0.6)(0.6)(800 \mathrm{keV})$; after three, it is $(0.6)^{3}(800 \mathrm{keV})$. Therefore, after $n$ collisions, the neutron energy is $(0.6)^{n}(800 \mathrm{keV})$. We want $n$ large enough so that
$$
(0.6)^{n}\left(8 \times 10^{5} \mathrm{eV}\right)=0.040 \mathrm{eV}
$$
Taking the logarithms of both sides of this equation yields
$$
\begin{array}{c}
n \log _{10} 0.6+\log _{10}\left(8 \times 10^{5}\right)=\log _{10} 0.04 \\
(n)(-0.222)+5.903=-1.398
\end{array}
$$
from which we find $n$ to be 32.9. So 33 collisions are required.

Km Neeraj
Km Neeraj
Numerade Educator
03:18

Problem 5

To examine the structure of a nucleus, pointlike particles with de Broglie wavelengths below about $10^{-16} \mathrm{~m}$ must be used. Through how large a potential difference must an electron fall to have this wavelength? Assume the electron is moving in a relativistic way.
The $\mathrm{KE}$ and momentum of the electron are related through
$$
\mathrm{KE}=\sqrt{p^{2} \mathrm{c}^{2}+m^{2} \mathrm{c}^{4}}-m \mathrm{c}^{2}
$$
Because the de Broglie wavelength is $\lambda=h / p$, this equation becomes
$$
\mathrm{KE}=\sqrt{\left(\frac{h \mathrm{c}}{\lambda}\right)^{2}+m^{2} \mathrm{c}^{4}}-m \mathrm{c}^{2}
$$
Using $\lambda=10^{-16} \mathrm{~m}, h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}$, and $m=9.1 \times 10^{-31} \mathrm{~kg}$, we find that
$$
\mathrm{KE}=1.99 \times 10^{-9} \mathrm{~J}=1.24 \times 10^{10} \mathrm{eV}
$$
The electron must be accelerated through a potential difference of about $10^{10} \mathrm{eV}$.

Km Neeraj
Km Neeraj
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08:35

Problem 6

The following fusion reaction takes place in the Sun and furnishes much of its energy:
$$
4{ }_{1}^{1} \mathrm{H} \rightarrow 4{ }_{2}^{4} \mathrm{He}+2_{+1}^{0} e+\text { energy }
$$
where $_{+1}^{0} e$ is a positron electron. How much energy is released as $1.00 \mathrm{~kg}$ of hydrogen is consumed? The masses of ${ }^{1} \mathrm{H},{ }^{4} \mathrm{He}$, and ${ }_{+}{ }^{0} e$ are, respectively, $1.007825,4.002604$, and $0.000549 \mathrm{u}$, where atomic
electrons are included in the first two values.

Ignoring the electron binding energy, the mass of the reactants, 4 protons, is 4 times the atomic mass of hydrogen $\left({ }^{1} \mathrm{H}\right)$, less the mass of 4 electrons:
$$
\begin{aligned}
\text { Reactant Mass } &=(4)(1.007825 \mathrm{u})-4 m_{e} \\
&=4.031300 \mathrm{u}-4 m_{e}
\end{aligned}
$$
where $m_{e}$ is the mass of the electron (or positron). The reaction products have a combined mass
$$
\begin{aligned}
\text { Product mass } &=\left(\text { Mass of }_{2}^{4} \text { He nucleus }\right)+2 m_{e} \\
&=\left(4.002604 \mathrm{u}-2 m_{e}\right)+2 m_{e} \\
&=4.002604 \mathrm{u}
\end{aligned}
$$
The mass loss is therefore
$$
(\text { Reactant mass })-(\text { Product mass })=\left(4.0313 \mathrm{u}-4 m_{e}\right)-4.0026 \mathrm{u}
$$
Substituting $m_{e}-0.000549$ u gives the mass loss as $0.0265 \mathrm{u}$.
But $1.00 \mathrm{~kg}$ of ${ }^{1} \mathrm{H}$ contains $6.02 \times 10^{26}$ atoms. For each four atoms that undergo fusion, $0.0265 \mathrm{u}$ is lost. The mass lost when $1.00 \mathrm{~kg}$ undergoes fusion is therefore
$$
\begin{aligned}
\text { Mass loss } / \mathrm{kg} &=(0.0265 \mathrm{u})\left(6.02 \times 10^{26} / 4\right)=3.99 \times 10^{24} \mathrm{u} \\
&=\left(3.99 \times 10^{24} \mathrm{u}\right)\left(1.66 \times 10^{-27} \mathrm{~kg} / \mathrm{u}\right)=0.00663 \mathrm{~kg}
\end{aligned}
$$
Then, from the Einstein relation.
$$
\Delta \mathrm{E}=(\Delta m) \mathrm{c}^{2}=(0.00663 \mathrm{~kg})\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)^{2}=5.96 \times 10^{14} \mathrm{~J}
$$

Abid Hussain
Abid Hussain
Numerade Educator
04:50

Problem 7

Lithium hydride, LiH, has been proposed as a possible nuclear fuel. The nuclei to be used and the reaction involved are as follows:
$$
\begin{array}{ccc}
{ }_{3}^{6} \mathrm{Li} \\
6.01513 & +{ }_{1}^{2} \mathrm{H} & & \rightarrow 2{ }_{2}^{4} \mathrm{He} \\
& 2.01410 & 4.00260
\end{array}
$$
the listed masses being those of the neutral atoms. Calculate the expected power production, in megawatts, associated with the consumption of $1.00 \mathrm{~g}$ of LiH per day. Assume 100 percent efficiency.
Ignoring the electron binding energies, the change in mass for the reaction must be computed first:
We find the loss in mass by subtracting the product mass from the reactant mass. In the process, the electron masses drop out and the mass loss is found to be $0.02403 \mathrm{u}$.

The fractional loss in mass is $0.0240 / 8.029=2.99 \times 10^{-3}$. Therefore, when $1.00 \mathrm{~g}$ reacts, the mass loss is
$$
\left(2.99 \times 10^{-3}\right)\left(1.00 \times 10^{-3} \mathrm{~kg}\right)=2.99 \times 10^{-6} \mathrm{~kg}
$$
This corresponds to an energy of
$$
\Delta \mathrm{E}=(\Delta m) \mathrm{c}^{2}=\left(2.99 \times 10^{-6} \mathrm{~kg}\right)\left(2.998 \times 10^{\mathrm{s}} \mathrm{m} / \mathrm{s}\right)^{2}=2.687 \times 10^{11} \mathrm{~J}
$$
Then
$$
\text { Power }=\frac{\text { Energy }}{\text { Time }}=\frac{2.687 \times 10^{11} \mathrm{~J}}{86400 \mathrm{~s}}=3.11 \mathrm{MW}
$$

Abid Hussain
Abid Hussain
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03:59

Problem 8

Cosmic rays bombard the $\mathrm{CO}_{2}$ in the atmosphere and, by nuclear reaction, cause the formation of the radioactive carbon isotope ${ }_{6}^{14} \mathrm{C}$. This isotope has a half-life of 5730 years. It mixes into the atmosphere uniformly and is taken up in plants as they grow. After a plant dies, the ${ }^{14} \mathrm{C}$ decays over the ensuing years. How old is a piece of wood that has a ${ }^{14} \mathrm{C}$ content which is only 9 percent as large as the average ${ }^{14} \mathrm{C}$ content of new-grown wood?
During the years, the ${ }^{14} \mathrm{C}$ has decayed to $0.090$ its original value. Hence (see Problem $45.6$ ),
$$
\frac{N}{N_{0}}=e^{-\lambda t} \quad \text { becomes } \quad 0.090=e^{-0.693 /(5730 \text { yeax })}
$$
After taking the natural logarithms of both sides,
$$
\begin{array}{c}
\ln 0.090=\frac{-0.693 t}{5730 \text { years }} \\
\text { from which } t=\left(\frac{5730 \text { years }}{-0.693}\right)(-2.41)=1.99 \times 10^{4} \text { years }
\end{array}
$$
The piece of wood is about 20000 years old.

Abid Hussain
Abid Hussain
Numerade Educator
04:32

Problem 9

Iodine-131 has a half-life of about $8.0$ days. When consumed in food, it localizes in the thyroid. Suppose $7.0$ percent of the ${ }^{131}$ I localizes in the thyroid and that 20 percent of its disintegrations are detected by counting the emitted gamma rays. How much ${ }^{131}$ I must be ingested to yield a thyroid count rate of 50 counts per second?
Because only 20 percent of the disintegrations are counted, there must be a total of $50 / 20 \%$ or $50 / 0.20=250$ disintegrations per second, which is what $\Delta N / \Delta t$ is. From Chapter 45 .
$$
\frac{\Delta N}{\Delta t}=\lambda N=\frac{0.693 N}{t_{1 / 2}} \quad \text { and so } \quad 250 \mathrm{~s}^{-1}=\frac{0.693 N}{(8.0 \mathrm{~d})(3600 \mathrm{~s} / \mathrm{h})(24 \mathrm{~h} / \mathrm{d})}
$$
from which $N=2.49 \times 10^{8}$.
However, this is only $7.0$ percent of the ingested ${ }^{131} \mathrm{I}$. Hence the number of ingested atoms is $N / 0.070=3.56 \times 10^{9}$. And, since $1.00 \mathrm{kmol}$ of ${ }^{131}$ i is approximately $131 \mathrm{~kg}$, this number of atoms represents
$$
\left(\frac{3.56 \times 10^{9} \text { atoms }}{6.02 \times 10^{26} \text { atoms } / \mathrm{kmol}}\right)(131 \mathrm{~kg} / \mathrm{kmol})=7.8 \times 10^{-16} \mathrm{~kg}
$$
which is the mass of ${ }^{131}$ I that must be ingested.

Km Neeraj
Km Neeraj
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04:55

Problem 10

A beam of gamma rays has a cross-sectional area of $2.0 \mathrm{~cm}^{2}$ and carries $7.0 \times 10^{8}$ photons through the cross section each second. Each photon has an energy of $1.25 \mathrm{MeV}$. The beam passes through a $0.75 \mathrm{~cm}$ thickness of flesh $\left(\rho=0.95 \mathrm{~g} / \mathrm{cm}^{3}\right)$ and loses $5.0$ percent of its intensity in the process. What is the average dose (in Gy and in rd) applied to the flesh each second?

The dose in this case is the energy absorbed per kilogram of flesh. Since $5.0 \%$ of the intensity is absorbed,
Number of photons absorbed/s $=\left(7.0 \times 10^{8} \mathrm{~s}^{-1}\right)(0.050)=3.5 \times 10^{7} \mathrm{~s}^{-1}$
and each such photon carries an energy of $1.25 \mathrm{MeV}$. Hence,
$$
\text { Energy absorbed } / \mathrm{s}=\left(3.5 \times 10^{7} \mathrm{~s}^{-1}\right)(1.25 \mathrm{MeV})=4.4 \times 10^{7} \mathrm{MeV} / \mathrm{s}
$$
We need the mass of flesh in which this energy was absorbed. The beam was delivered to a region of area $2.0$ $\mathrm{cm}^{2}$ and thickness $0.75 \mathrm{~cm}$. Thus,
$$
\text { Mass }=\rho V=\left(0.95 \mathrm{~g} / \mathrm{cm}^{3}\right)\left[\left(2.0 \mathrm{~cm}^{2}\right)(0.75 \mathrm{~cm})\right]=1.43 \mathrm{~g}
$$
Keeping in mind that $1 \mathrm{rd}=0.01 \mathrm{~Gy}$,
$$
\text { Dose/s }=\frac{\text { Energy/s }}{\text { Mass }}=\frac{\left(4.4 \times 10^{7} \mathrm{MeV} / \mathrm{s}\right)\left(1.6 \times 10^{-13} \mathrm{~J} / \mathrm{MeV}\right)}{1.43 \times 10^{-3} \mathrm{~kg}}=4.9 \mathrm{mGy} / \mathrm{s}=0.49 \mathrm{rd} / \mathrm{s}
$$

Abid Hussain
Abid Hussain
Numerade Educator
02:54

Problem 11

A beam of alpha particles passes through flesh and deposits $0.20 \mathrm{~J}$ of energy in each kilogram of flesh. The $Q$ for these particles is $12 \mathrm{~Sv} / \mathrm{Gy}$. Find the dose in $\mathrm{Gy}$ and $\mathrm{rd}$, as well as the effective dose in Sv and rem.
Recall that $H=Q D$ where
$$
D=\text { Dose }=\frac{\text { Absorbed energy }}{\text { Mass }}=0.20 \mathrm{~J} / \mathrm{kg}=0.20 \mathrm{~Gy}=20 \mathrm{rd}
$$
Hence, $H=$ Effective dose $=Q($ dose $)=(12 \mathrm{~Sv} / \mathrm{Gy})(0.20 \mathrm{~Gy})=2.4 \mathrm{~Sv}=2.4 \times 10^{2} \mathrm{rem}$

Abid Hussain
Abid Hussain
Numerade Educator
05:08

Problem 12

A tumor on a person's leg has a mass of $3.0 \mathrm{~g} .$ What is the minimum activity a radiation source can have if it is to furnish a dose of $10 \mathrm{~Gy}$ to the tumor in $14 \mathrm{~min}$ ? Assume each disintegration within the source, on the average, provides an energy $0.70 \mathrm{MeV}$ to the tumor.
A dose of 10 Gy corresponds to $10 \mathrm{~J}$ of radiation energy being deposited per kilogram. Since the tumor has a mass of $0.0030 \mathrm{~kg}$, the energy required for a 10 Gy dose is $(0.0030 \mathrm{~kg})(10 \mathrm{~J} / \mathrm{kg})=0.030 \mathrm{~J}$.
Each disintegration provides $0.70 \mathrm{MeV}$, which in joules is
$$
\left(0.70 \times 10^{6} \mathrm{eV}\right)\left(1.60 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)=1.12 \times 10^{-13} \mathrm{~J}
$$
A dose of 10 Gy requires that an energy of $0.030 \mathrm{~J}$ be delivered. That total energy divided by the energy per disintegration, yields the number of disintegrations:
$$
\frac{0.030 \mathrm{~J}}{1.12 \times 10^{-13} \mathrm{~J} / \text { disintegration }}=2.68 \times 10^{11} \text { disintegrations }
$$
They are to occur in $14 \mathrm{~min}$ (or $840 \mathrm{~s}$ ), and so the disintegration rate is
$$
\frac{2.68 \times 10^{11}}{840 \mathrm{~s}} \text { disintegrations }=3.2 \times 10^{8} \text { disintegrations/s. }
$$
Hence, the source activity must be at least $3.2 \times 10^{8} \mathrm{~Bq}$. Since $1 \mathrm{Ci}=3.70 \times 10^{10} \mathrm{~Bq}$, the source activity must be at least $8.6 \mathrm{mCi}$.

Abid Hussain
Abid Hussain
Numerade Educator
06:30

Problem 13

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 $)=(14)(188)=2.6 \times 10^{4} \mathrm{~Sv}$

Abid Hussain
Abid Hussain
Numerade Educator
03:46

Problem 14

Consider the following fission reaction:
$$
\begin{array}{cccccc}
{ }_{0}^{1} n & +{ }_{92}^{235} \mathrm{U} & \rightarrow{ }_{56}^{138} \mathrm{Ba} & +{ }_{41}^{93} \mathrm{Nb} & +{ }_{0}^{1} n & + & 5_{-1}^{0} e \\
1.0087 & 235.0439 & 137.9050 & 92.9060 & 1.0087 & 0.00055
\end{array}
$$
where the neutral atomic masses are given. How much energy is released when $(a) 1$ atom undergoes this type of fission, and $(b) 1.0 \mathrm{~kg}$ of atoms undergoes fission?

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
05:16

Problem 15

It is proposed to use the nuclear fusion reaction
$$
\begin{array}{rr}
2{ }_{1}^{2} \mathrm{H} & \rightarrow{ }_{2}^{4} \mathrm{He} \\
2.014102 & 4.002604
\end{array}
$$
to produce industrial power (neutral atomic masses are given). If the output is to be $150 \mathrm{MW}$ and the energy of the reaction will be used with 30 percent efficiency, how many grams of deuterium fuel will be needed per day?

Khoobchandra Agrawal
Khoobchandra Agrawal
Numerade Educator
03:01

Problem 16

One of the most promising fusion reactions for power generation involves deuterium $\left({ }^{2} \mathrm{H}\right)$ and tritium $\left({ }^{3} \mathrm{H}\right)$ :
$$
\begin{array}{cccc}
{ }_{1}^{2} \mathrm{H} & +{ }_{1}{ }_{1}^{3} \mathrm{H} & \rightarrow & { }_{2}^{4} \mathrm{He} & +\underset{0}{1}{ }_{0}^{1} n \\
2.01410 & 3.01605 & 4.00260 & 1.00867
\end{array}
$$
where the atomic masses including electrons are as given. How much energy is produced when $2.0 \mathrm{~kg}$ of ${ }^{2} \mathrm{H}$ fuses with $3.0 \mathrm{~kg}$ of ${ }^{3} \mathrm{H}$ to form ${ }^{4} \mathrm{He}$ ?
46.17 [I] What is the average KE of a neutron at the center of the Sun, where the temperature is about $10^{7} \mathrm{~K}$ ? Give your answer to two significant figures.

Km Neeraj
Km Neeraj
Numerade Educator
01:52

Problem 17

What is the average KE of a neutron at the center of the Sun, where the temperature is about $10^{7} \mathrm{~K}$ ? Give your answer to two significant figures.

Km Neeraj
Km Neeraj
Numerade Educator
01:46

Problem 18

Find the energy released when two deuterons $\left({ }_{1}^{2} \mathrm{H}\right.$, atomic mass $=2.01410 \mathrm{u}$ ) fuse to form ${ }_{2}^{2} \mathrm{He}$ (atomic mass $=3.01603 \mathrm{u}$ ) with the release of a neutron. Give your answer to three significant figures.

Km Neeraj
Km Neeraj
Numerade Educator
01:51

Problem 19

The tar in an ancient tar pit has a ${ }^{14} \mathrm{C}$ activity that is only about $4.00$ percent of that found for new wood of the same density. What is the approximate age of the tar?

Km Neeraj
Km Neeraj
Numerade Educator
02:53

Problem 20

Rubidium- 87 has a half-life of $4.9 \times 10^{10}$ years and decays to strontium- 87, which is stable. In an ancient rock, the ratio of ${ }^{87} \mathrm{Sr}$ to ${ }^{87} \mathrm{Rb}$ is $0.0050 .$ If we assume all the strontium came from rubidium decay, about how old is the rock? Repeat if the ratio is $0.210$.

Km Neeraj
Km Neeraj
Numerade Educator
03:34

Problem 21

The luminous dial of an old watch gives off 130 fast electrons each minute. Assume that each electron has an energy of $0.50 \mathrm{MeV}$ and deposits that energy in a volume of skin that is $2.0 \mathrm{~cm}^{2}$ in area and $0.20 \mathrm{~cm}$ thick. Find the dose (in both Gy and rd) that the volume experiences in $1.0$ day. Take the density of skin to be $900 \mathrm{~kg} / \mathrm{m}^{3}$

Nicholas Majtenyi
Nicholas Majtenyi
Numerade Educator
02:54

Problem 22

An alpha-particle beam enters a charge collector and is measured to carry $2.0 \times 10^{-14} \mathrm{C}$ of charge into the collector each second. The beam has a cross-sectional area of $150 \mathrm{~mm}^{2}$, and it penetrates human skin to a depth of $0.14 \mathrm{~mm}$. Each particle has an initial energy of $4.0 \mathrm{MeV}$. The $Q$ for such particles is about 15 . What effective dose, in $\mathrm{Sv}$ and in rem, does a person's skin receive when exposed to this beam for $20 \mathrm{~s}$ ? Take $\rho=900 \mathrm{~kg} / \mathrm{m}^{3}$ for skin.

Abid Hussain
Abid Hussain
Numerade Educator