Answer:
The energy in its ground state is 10 meV.
Step-by-step explanation:
It is given that,
The energy of the electron in its first excited state is 40 meV.
Energy of the electron in any state is given by :
![E=(n^2\pi^2h^2)/(8mL^2)](https://img.qammunity.org/2020/formulas/physics/college/jteixch59guqr8kb6wf4x513wrbo09dq9o.png)
For ground state, n = 1
.............(1)
For first excited state, n = 2
.............(2)
Dividing equation (1) and (2), we get :
![(E_1)/(40)=(1)/(4)](https://img.qammunity.org/2020/formulas/physics/college/wc980lu4rs8l4kunu0j8peijwism3rahvp.png)
![E_1=10\ meV](https://img.qammunity.org/2020/formulas/physics/college/mi3f4rj6un17qz5hwt1p5oyy2gdzwiac58.png)
So, the energy in its ground state is 10 meV. Hence, this is the required solution.