Scrodinger equation: infinite square well problem.

[Benzoate]

Homework Statement

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

Homework Equations

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

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?

 

[Benzoate]

someone please respond to my problem. Anybody. Does my problem appear unreadable?

 

[ogb p]

The Schrodinger equation is a fundamental equation in quantum mechanics that describes the behavior and evolution of quantum systems. In this case, the infinite square well problem is a simplified model used to study the behavior of particles confined within a potential well.

In order to solve for the ground state energy of the electron and proton in the nucleus, we can use the formula E(n) = n^2 * (h/(2*pi))^2 / (2*m*L^2). Using the given values, we can calculate the ground state energy for the electron as 6e-8 joules and for the proton as 3.28e-11 joules.

For part c, we can calculate the energy difference between the ground state and the first excited state by using the same formula and substituting n=2 for the first excited state. This gives us a difference of 3.24e-11 joules for the electron and 1.31e-10 joules for the proton.

It is important to note that this calculation is based on a simplified model and does not take into account the complexities of the actual nuclear structure. Additionally, the discovery of the neutron has shown that the nucleus contains more particles than just electrons and protons. Nonetheless, this exercise provides a useful example of how the Schrodinger equation can be applied to solve for the energy levels of particles in a confined potential well.