Question

An electron istrapped in a finite well. How "far" (in eV) is it from being free if the penetration length of its wave function into the classically forbidden region $\mathbf{1}\mathbf{}\mathbf{\text{nm}}$?

Short answer

The distance of the electron to set free is $0.038\text{eV.}$

Step-by-step solution

Identification of given data

The penetration length of the classically forbidden region is$L=1\text{nm}$.

Concept/Significance of a potential well

A potential well is a place in aforce field, where the atomic nucleus is located and where the potential is much lower than at a point immediately outside it unless aparticle accumulates a significant amount of energy.

Determination of the distance of the electron from being free

The distance of the electron is mathematically presented as:

$$\begin{gathered} \delta =\frac{\hslash }{\sqrt{2m\left(U_0-E\right)}} \\ {U}_0-E=\frac{2m{\delta }^2}{{\hslash }^2} \end{gathered}$$

Here, $\hslash$is the reduce planks constant whose value is $1.055\times {10}^{-34}\text{J}·\text{s}$ , is the mass of the electron whose value is $9.11\times {10}^{-31}\text{kg}$ , and $\left(U_0-E\right)$is the distance of the potential well to free the electron.

Replaceall the values in the above equation:

$\begin{array}{rcl}{U}_0-E& =& \frac{\left(2\times 9.11\times {10}^{-31}\text{kg}\right){\left({10}^{-9}\text{m}\right)}^2}{{\left(1.055\times {10}^{-34}\text{J}·\text{s}\right)}^2}\\ & =& 6.1\times {10}^{-21}\text{J}\left(\frac{6.24\times {10}^{18}\text{eV}}{1\text{J}}\right)\\ & =& 38.064\times {10}^{-3}\text{eV}\\ & =& 0.038\text{eV.}\\ & & \end{array}$

Hence, the distance of the electron to set free is $0.038\text{eV.}$