Question 41.6: The Penetration Distance of Particle in a Finite Square Well......
The Penetration Distance of Particle in a Finite Square Well
When a particle in a finite square well is in its ground state, its penetration distance is Λ_1. The well’s height is U_0 = 5.55 eV. When the same system is in an excited state with an energy of 5.35 eV, its penetration distance is 4Λ_1. What is the system’s ground-state energy (in eV)?
INTERPRET and ANTICIPATE
The penetration distance depends on the system’s energy (Eq. 41.21).
We expect the ground-state energy to be lower than the excited state’s energy.
SOLVE
We are given that when the system is in the excited state, the penetration distance is four times longer than when it is in the ground state.
Substitute Equation 41.21 into Equation (1) and simplify.
\begin{aligned}& \frac{h}{2 \pi \sqrt{2 m\left(U_0-E_x\right)}}=\frac{4 h}{2 \pi \sqrt{2 m\left(U_0-E_1\right)}} \\ & \frac{1}{\sqrt{\left(U_0-E_x\right)}}=\frac{4}{\sqrt{\left(U_0-E_1\right)}}\end{aligned}Solve for E_1.
\begin{aligned}& E_1=16 E_x-15 U_0 \\& E_1=16(5.35 eV )-15(5.55 eV ) \\& E_1=2.35 eV\end{aligned}CHECK and THINK
As expected, the ground-state energy is less than the excited state’s energy.