Question

Is the ground-state energy of a proton trapped in a one-dimensional infinite potential well greater than, less than, or equal to that of an electron trapped in the same potential well?

Short answer

The ground state energy of a proton that trapped in the one dimensional potential well is less than the ground state energy of an electron that trapped in the same one dimensional potential well.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

The quantum number for ground state is, n = 1.

Significance of electron in an infinite potential well

The value of the energy of a charged particle that trapped in one dimensional infinite potential well of specific length depends on the value of the Plank’s constant, well length and mass of the charged particle.

Determination of the ground state energy of an electron that trapped in a one dimensional infinite potential well

The expression to calculate the ground state energy of an electron that trapped in a one dimensional potential well is expressed as,

${\left(E_n\right)}_e=\left(\frac{h^2}{8m_e{L}^2}\right)n^2$

Here, ${\left(E_n\right)}_e$is the ground state energy of an electron that trapped in a one dimensional potential well, h is the Plank’s constant whose value is $6.63\times {10}^{-34}J.s$, $m_e$is the mass an electron whose value is $9.11\times {10}^{-31}kg$and L is the length of the well.

Substitute all the known values in the above equation.

role="math" localid="1661773209232" $$\begin{gathered} {\left(E_1\right)}_e=\left[\frac{{\left(6.63\times {10}^{-34}J.s\right)}^2}{8\left(9.11\times {10}^{-31}kg\right)L^2}\right]{\left(1\right)}^2 \\ \approx \left[\frac{1}{L^2}\left(6.03\times {10}^{-38}\right)\right]J^2.s^2/kg \end{gathered}$$

Determination of the ground state energy of a proton that trapped in the same one dimensional infinite potential well

The expression to calculate the ground state energy of a proton that trapped in the same one dimensional potential well is expressed as,

${\left(E_n\right)}_p=\left(\frac{h^2}{8m_p{L}^2}\right)n^2$

Here,${\left(E_n\right)}_p$ is the ground state energy of a proton that trapped in the same one dimensional potential well,$m_p$ is the mass a proton whose value is$1.67\times {10}^{-27}kg$ and L is the length of the well.

Substitute all the known values in the above equation.

$$\begin{gathered} {\left(E_n\right)}_p=\left[\frac{{\left(6.63\times {10}^{-34}J.s\right)}^2}{8\left(1.67\times {10}^{-27}kg\right)L^2}\right]{\left(1\right)}^2 \\ \approx \left[\frac{1}{L^2}\left(3.29\times {10}^{-41}\right)\right]J^2.s^2/kg \end{gathered}$$

From the above calculation, one can observe that the ground state energy of a proton that trapped in the one-dimensional potential well is$\left[\frac{1}{L^2}\left(3.29\times {10}^{-41}\right)\right]J^2.s^2/kg$ that is less than the ground state energy of an electron that trapped in the same one-dimensional potential well$\left[\frac{1}{L^2}\left(6.03\times {10}^{-38}\right)\right]J^2.s^2/kg$ .