Respuesta :
Answer:
The energy in its ground state is 10 meV.
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 :
[tex]E=\dfrac{n^2\pi^2h^2}{8mL^2}[/tex]
For ground state, n = 1
[tex]E_1=\dfrac{\pi^2h^2}{8mL^2}[/tex].............(1)
For first excited state, n = 2
[tex]40=\dfrac{2^2\pi^2h^2}{8mL^2}[/tex].............(2)
Dividing equation (1) and (2), we get :
[tex]\dfrac{E_1}{40}=\dfrac{1}{4}[/tex]
[tex]E_1=10\ meV[/tex]
So, the energy in its ground state is 10 meV. Hence, this is the required solution.
