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Math Help - Math/Science twist

  1. #1
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    Math/Science twist

    I found this question a bit confusing. Anyone who can help out with these answers.Your help is appreciated

    In a main sequence star such as our Sun, a fusion reaction is ongoing that releases energy: the
    pp chain, in which four hydrogen-1 nuclei (single protons) become one helium nucleus (alpha
    particle). The resulting mass is less than the initial mass: this ‘mass loss’ is the source of Sun’s
    energy which can be calculated from the equation: E= m*c
    2 , where E is the energy released;
    m is the ‘missing’ mass; and c is the speed of light,
    3.0 * 105 km/s. The mass of one proton is
    1.008 a.m.u (atoms mass unit: the mass of a carbon-12 nucleus is defined to be 12 a..m.u). The
    mass of a helium-4 nucleus is 4.003 a.m.u: there is a deficit. 1 a.m.u = 1.66*10
    -27 kg and it is
    estimated that 3.6
    ×1038 protons are converted into helium nuclei every second.

    A. What is the energy released in a single reaction where four protons ->
    one He nucleus?

    B. What is the power output of Sun? (3 marks)
    Power = (number of reactions per second) X (energy released per reaction)

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  2. #2
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    Quote Originally Posted by srr2612 View Post
    I found this question a bit confusing. Anyone who can help out with these answers.Your help is appreciated

    In a main sequence star such as our Sun, a fusion reaction is ongoing that releases energy: the
    pp chain, in which four hydrogen-1 nuclei (single protons) become one helium nucleus (alpha
    particle). The resulting mass is less than the initial mass: this ‘mass loss’ is the source of Sun’s
    energy which can be calculated from the equation: E= m*c
    2 , where E is the energy released;
    m is the ‘missing’ mass; and c is the speed of light,
    3.0 * 105 km/s. The mass of one proton is
    1.008 a.m.u (atoms mass unit: the mass of a carbon-12 nucleus is defined to be 12 a..m.u). The
    mass of a helium-4 nucleus is 4.003 a.m.u: there is a deficit. 1 a.m.u = 1.66*10
    -27 kg and it is
    estimated that 3.6
    ×1038 protons are converted into helium nuclei every second.

    A. What is the energy released in a single reaction where four protons ->
    one He nucleus?

    B. What is the power output of Sun? (3 marks)
    Power = (number of reactions per second) X (energy released per reaction)

    Something like the following:

    Given: Mass of Hydrogen-1= 1.008 amu, mass of Helium-4=4.003 amu, four hydrogen-1 become one helium-4, speed of light = 3.0(10^5)km/s (we'll need this in m/s so converting 3.0(10^8)m/s), and 1 amu= 1.66(10^{-27})kg.

    Take 4 times the mass of hydrogen, 4.032 amu, and subtract from it the helium mass, 4.003 amu, to get \Delta m=0.029amu this is also our missing mass to soon be used after we convert to kg.

    \Delta m=0.029 amu|\frac{1.66(10^{-27})kg}{1 amu}
    After multiplying across and canceling the amu we're free to use E=mc^2

    c^2 = (3.0(10^8)m/s)^2 = 9.0(10^16)m^2/s^2

    E = (4.814^{-29}kg)(9.0(10^{16})m^2/s^2) = 4.3326(10^{-12})J

    A) Keeping significant digits the energy is about 4.3(10^{-12})J

    for the second question we need to know how many helium are formed per second and we are given that 3.6(10^{38})p^{+} are converted to helium every second so divide that number by four, the number of proton it takes to create one helium, 9(10^{37}) reactions per second.

    Power = (number of reactions per second) X (energy released per reaction)
    Power = 9(10^{37})He/s X 4.3326(10^{-12})J
    Power = 3.89934(10^{26})J/s

    B) Power of the sun is about 3.9(10^{26})J/s
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