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Scrodinger equation: infinite square well problem.

  1. Oct 22, 2007 #1
    1. The problem statement, all variables and given/known data

    In the early days of nuclear physics before the neutron was discovered , it was thought that the nucleus contained only electrons and protons. If we consider the nucleus to be a one-dimensional infinite well with L=1e-15 m and ignore relativity, compute the ground state energy for a) the electron and b) the proton in the nucleus. c) compute the energy difference between the ground state and the first excited state. for each particle

    2. Relevant equations

    E(n)=n^2 *(h/(2*pi))^2/(2*m*L^2)

    3. The attempt at a solution

    part a and b wasn't very difficult. a)E=n^2 *(h/(2*pi))^2/(2*m*L^2)=pi^2*(1.054e-34 J*s)^2/((2)*(9.11e-31 kg)(1e-15)^2=6e-8 joules and b) E=n^2 *(h/(2*pi))^2/(2*m*L^2)=pi^2*(1.054e-34 J*s)^2/((2)*(1.67e-27 kg)(1e-15)^2 =3.28e-11 joules. I had trouble with part c. should I assume the first excited state is n=2?
  2. jcsd
  3. Oct 22, 2007 #2
    someone please respond to my problem. Anybody. Does my problem appear unreadable?
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