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?