Question

A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an energy of 6.0 eV. What is the energy of this particle in the ground state?

Answer 1

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy $E_{n}$ of a particle of mass m in the nth energy state of an infinite square well potential with width L is:

                                                    $E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}$

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

                                                    $E_{1}=\frac{h^{2}}{8mL^{2}}$            

                                                    $E_{2}=\frac{h^{2}}{2mL^{2}}$

So we can rewrite the energy in the ground state as:

                                                   $E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})$

                                                      $E_{1}=\frac{1}{4} E_{2}$

                                                   $E_{1}=\frac{1}{4} ( 6.0\ eV)$

Finally

                                                    $E_{1}=1.5\ eV$