The probability of nuclear fusion is greatly enhanced when the appropriate nuclei are brought close together, but their mutual coulomb repul

Question

The probability of nuclear fusion is greatly enhanced when the appropriate nuclei are brought close together, but their mutual coulomb repulsion must be overcome. This can be done using the kinetic energy of high temperature gas ions or by accelerating the nuclei toward one another.

Required:
a. Calculate the potential energy of two singly charged nuclei separated by 1.00×10^−12m
b. At what temperature will atoms of a gas have an average kinetic energy equal to this needed electrical potential energy?

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Thiên Hương 3 years 2021-08-23T04:20:22+00:00 1 Answers 12 views 0

Answers ( )

  1. kimcuc
    0
    2021-08-23T04:21:27+00:00
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    Answer:

    a

      PE = 2.3 *10^{-16} \ J

    b

     T = 1.1 *10^{7} \ K

    Explanation:

    From the question we are told that

          The distance of separation is  d = 1.00 *10^{-12} \ m

    Generally the electric potential energy can be mathematically represented as

                PE = \frac{k * q_1 q_2 }{d}

    Given that in a nuclei the only charged particle is the proton who charge is

         p = 1.60 *10^{-19} \ C

    Hence

         q_1 = q_2 = 1.60 *10 ^{-19} \ C

    And k is the coulomb constant with values   k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.N/A2

          So we have that

           PE = \frac{9*10^9 * (1.60 *10^{-19})^2}{ 1.00*10^{-12}}

          PE = 2.3 *10^{-16} \ J

    The relationship between the electrical potential energy and the temperature is mathematically represented as

             PE = \frac{3}{2} kT

    Here  k is  the Boltzmann’s constant with value  k = 1.38*10^{-23} JK^{-1}

       making T the subject

           T = \frac{2}{3} * \frac{PE}{k}

    substituting values

          T = \frac{2}{3} * \frac{2.30 *10^{-16}}{ 1.38 *10^{-23}}

         T = 1.1 *10^{7} \ K

     

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