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Math Help - Anyone good with Chemistry?? Some nuclear chemistry questions

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    Anyone good with Chemistry?? Some nuclear chemistry questions

    Question 1 1 points Save
    An atom with 15 protons and 18 neutrons would be an isotope of

    argon.
    sulfur.
    arsenic.
    phosphorus.
    chlorine.

    Question 2 1 points Save
    Relative to the sum of the masses of its constituent nucleons (neutrons + protons), the mass of a nucleus is

    always the same.
    always greater.
    always smaller.
    sometimes the same and sometimes smaller.
    sometimes greater and sometimes smaller.

    Question 3 1 points Save
    WhenU is bombarded with one neutron, fission occurs and the products are three neutrons,Cs and

    Rb.
    Cs.
    Rb.
    Rb.
    Ba.

    Question 4 1 points Save
    The half-life of the radioisotope Eu is 1.0 hr. The decay of a 160.0-g sample of the isotope to 1.25 g requires

    3.0 hr.
    4.0 hr.
    5.0 hr.
    6.0 hr.
    7.0 hr.

    Question 5 1 points Save
    A living tree contains C (half-life 5,600 years) and has a specific activity of 750 counts per hour. A wooden artifact recovered from an archeological site gives a count of 190 counts per hour. The age of this artifact is most nearly

    5,600 years.
    11,000 years.
    17,000 years.
    22,000 years.
    47,000 years.

    Question 6 1 points Save
    The half-life of K is 12.5 hours. How much will remain after 100 hours if the original sample contained 256 g of K?

    16 g
    8 g
    4 g
    2 g
    1 g

    Question 7 1 points Save
    When Cu undergoes positron emission, what is the immediate nuclear product?

    Zn
    Cu
    Ni
    Zn
    Ni

    Question 8 1 points Save
    As a radioactive isotope decays, its rate constant

    decreases.
    halves.
    increases.
    doubles.
    remains the same.

    Question 9 1 points Save
    The isotope Pb has a half-life of 22 years. What percentage of a pure Pb sample prepared in April 1937 remains in April 1993?

    log = ; t1/2 =

    38%
    31%
    26%
    21%
    17%

    Question 10 1 points Save
    What is the binding energy per nucleon, in joules, of a chlorine-35 atom, given that the atomic mass equals 34.9595 amu?

    (e- mass = 0.00055 amu, n mass = 1.00867 amu, p mass = 1.00728 amu,

    c = 3.00 x 108 m/s, 1 amu = 1.6606 x 10-24 g, 1 J = 1 kg m2/s2)

    1.56 x 10-12 J/nucleon
    1.56 x 10-11 J/nucleon
    1.56 x 10-10 J/nucleon
    1.41 x 10-12 J/nucleon
    4.93 x 10-11 J/nucleon
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Stuck686 View Post
    Question 1 1 points Save
    An atom with 15 protons and 18 neutrons would be an isotope of

    argon.
    sulfur.
    arsenic.
    phosphorus.
    chlorine.
    The non-isotopic (that is to say, the most common) numbers for the answers are:
    Ar ==> 18 p, 22 n
    S ==> 16 p, 16 n
    As ==> 33 p, 42 n
    P ==> 15 p, 16 n
    Cl ==> 17 p, 18 n

    An isotope is the same element, but with a different atomic mass. So our nucleus needs 15 p. Thus the answer is P.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Stuck686 View Post
    Question 2 1 points Save
    Relative to the sum of the masses of its constituent nucleons (neutrons + protons), the mass of a nucleus is

    always the same.
    always greater.
    always smaller.
    sometimes the same and sometimes smaller.
    sometimes greater and sometimes smaller.
    Sometimes the same and sometimes smaller.

    Why you ask? When you put several protons and neutrons together they have to overcome an electrostatic repulsion. To make a long story short, there has to be some attractive force in the nucleus that dominates the electrostatic interaction: aka the nucleus sits in a potential energy well. The difference between the mass energy of the constituents and the mass energy of the nucleus is called the "binding energy" of the nucleus.

    So how can we say "sometimes the same?" Consider a Hydrogen atom. It has one proton and nothing else. So the mass of the nucleus had better be the same as the mass of the constituent! (Technically this isn't true, since in Hydrogen the proton is in a stable configuration with an electron. This should create a small mass difference, but I don't know if it would be measurable.)

    -Dan
    Last edited by topsquark; April 28th 2007 at 04:01 PM.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Stuck686 View Post
    Question 3 1 points Save
    WhenU is bombarded with one neutron, fission occurs and the products are three neutrons,Cs and

    Rb.
    Cs.
    Rb.
    Rb.
    Ba.
    I'm assuming you are starting with U238? And you have:
    U + n --> 3n + Cs + ?
    as a reaction?

    Unless a beta or positron decay occurs (and we know it doesn't since neither is listed among the products) then all you need to do is "balance" the number of protons and neutrons before and after the reaction. Let the unknown product X have x p and y n:
    Before: U + n: 92 p, 147 n
    After: 55 + x p, 3*1 + 78 + y

    So
    x + 55 = 92 ==> x = 37
    y + 81 = 147 ==> y = 66

    So the product looks like its Rb103.

    -Dan
    Last edited by topsquark; April 30th 2007 at 05:32 AM.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Stuck686 View Post
    Question 4 1 points Save
    The half-life of the radioisotope Eu is 1.0 hr. The decay of a 160.0-g sample of the isotope to 1.25 g requires

    3.0 hr.
    4.0 hr.
    5.0 hr.
    6.0 hr.
    7.0 hr.
    The equation is:
    m(t) = m(0)*e^{-(lambda)*t}
    where (lambda) is a constant.

    We know that the half-life is 1.0 h, so we know that
    m(1) = (1/2)*m(0)
    m(1) = m(0)*e^{-(lambda)*1}

    (1/2)m(0) = m(0)*e^{-(lambda)}

    1/2 = e^{-(lambda)}

    -(lambda) = ln(1/2) = -ln(2)

    (lambda) = ln(2) = 0.6931 (or so.)

    Thus
    m(t) = m(0)*e^{-0.6931*t}

    We have that m(t) = 1.25 g and m(0) = 160.0 g thus:
    1.25 = 160*e^{-0.6931*t}

    e^{-0.6931*t} = 1.25/160 = 0.0078125

    -0.6931*t = ln(0.0078125) = -4.8520

    t = 7 hours

    -Dan

    (Note: You can do a similar problem using:
    m(t) = m(0)*2^{-(lambda)t}
    The end result is the same.)
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    Senior Member ecMathGeek's Avatar
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    topsquark, I wish I knew about this site when I was in Calculus based Physics. You're awesome.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Stuck686 View Post
    Question 5 1 points Save
    A living tree contains C (half-life 5,600 years) and has a specific activity of 750 counts per hour. A wooden artifact recovered from an archeological site gives a count of 190 counts per hour. The age of this artifact is most nearly

    5,600 years.
    11,000 years.
    17,000 years.
    22,000 years.
    47,000 years.
    Hmmm... It's been a while for this so I might "argue a few circles" before I get to the answer.

    The specific activity will be directly proportional to the amount of radioactive material present. Thus if we take the ratio of the specific activity of the sample vs the specific activity of a living sample we should get the ratio of the present mass (m(t)) vs the original mass (m(0)) that was present in the sample when it was alive.

    But
    m(t)/m(0) = e^{-(lambda)t}
    according to the decay equation.

    We can easily get that (lambda) = 0.0001238 or so.

    Thus
    190/750 = m(t)/m(0) = e^{-0.0001238*t}

    and I get that
    t = 11093 years

    So around 11000 years.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Stuck686 View Post
    Question 6 1 points Save
    The half-life of K is 12.5 hours. How much will remain after 100 hours if the original sample contained 256 g of K?

    16 g
    8 g
    4 g
    2 g
    1 g
    You should be able to do this one yourself now. I got 1 g.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Stuck686 View Post
    Question 7 1 points Save
    When Cu undergoes positron emission, what is the immediate nuclear product?

    Zn
    Cu
    Ni
    Zn
    Ni
    We have the reaction:
    Cu --> e+ + ?
    (where e+ is the positron.)

    The process by which this happens is that a positron (the antimatter version of the electron) decays from a proton:
    p --> n + e+ + (electron neutrino)

    So as you can see we are reducing the atomic number by 1 unit and not changing the nuclear mass (by any appreciable amount anyway.) Cu has an atomic number of 29, so our prduct has an atomic numer of 28, aka Ni.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    topsquark, I wish I knew about this site when I was in Calculus based Physics. You're awesome.
    Gracias senor!

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Stuck686 View Post
    Question 9 1 points Save
    The isotope Pb has a half-life of 22 years. What percentage of a pure Pb sample prepared in April 1937 remains in April 1993?

    log = ; t1/2 =

    38%
    31%
    26%
    21%
    17%
    You should be able to do this one now as well. I got 17%.

    Quote Originally Posted by Stuck686 View Post
    Question 10 1 points Save
    What is the binding energy per nucleon, in joules, of a chlorine-35 atom, given that the atomic mass equals 34.9595 amu?

    (e- mass = 0.00055 amu, n mass = 1.00867 amu, p mass = 1.00728 amu,

    c = 3.00 x 108 m/s, 1 amu = 1.6606 x 10-24 g, 1 J = 1 kg m2/s2)

    1.56 x 10-12 J/nucleon
    1.56 x 10-11 J/nucleon
    1.56 x 10-10 J/nucleon
    1.41 x 10-12 J/nucleon
    4.93 x 10-11 J/nucleon
    This is pretty much just unit conversion. The atomic mass is 34.9595 amu meaning we have 35 nucleons in the nucleus. We have 17 p, 18 n in a Chlorine nucleus, so:
    17*(1.00728 amu) + 18*(1.00867 amu) + 35*(0.00055 amu) - 34.9595 amu = 0.33957 amu

    Now for the unit conversions:
    m = 0.33957 amu *(1.6606 x 10^{-24} g)/(1 amu) = 5.6389 x 10^{-25} g = 5.6389 x 10^{-28} kg
    (This is called the "binding mass" by the way.)

    What is this in Joules? Use Einstein's famous (rest) mass to energy converter:
    E = mc^2

    E = (5.6389 x 10^{-28} kg)*(3 x 10^8 m/s)^2 = 5.07501 x 10^{-11} J

    We aren't done! This is the binding energy of the nulceus. To get the binding energy per nucleon, divide by 35:
    e = E/35 = 1.45 x 10^{-12} J/nucleon
    (I presume this corresponds to the 1.41 x 10^{-12} J/nucleon answer.)

    By the way, this is a tiny energy, but in particle physics terms its big. The binding energy per nucleon here corresponds to an energy of 9 MeV/nucleon. (Essentially representing the mass-energy of 18 electrons for each nucleon. So technically we should have considered the kinetic energy of the electrons as well.)

    -Dan
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