Respuesta :
Answer:
[tex]E'_n=4\times E_n[/tex]
Explanation:
The energy of a particle in one dimensional box is given by :
[tex]E_n=\dfrac{n^2h^2}{8mL^2}[/tex]
n is the quantum number
h is the Planck's constant
m is the mass of particle
L is the length of box
If L' = L/2
Energy level is given by :
[tex]E'_n=\dfrac{n^2h^2}{8mL'^2}[/tex]
[tex]E'_n=\dfrac{n^2h^2}{8m(L/2)^2}[/tex]
[tex]E'_n=4\times \dfrac{n^2h^2}{8mL^2}[/tex]
[tex]E'_n=4\times E_n[/tex]
So, if the same particle is placed in another box of length L/2, the energy should be four times the energy for the first box. Hence, this is the required solution.
