Question

A certain atom requires 3.0 eV of energy to excite an electron from the ground level to the first excited level. Model the atom as an electron in a box and find the width L of the box.

Short answer

The width L of the box is $6.14\times {10}^{-10}m$.

Step-by-step solution

Define the energy level.

The energy level of a particle of massin a box of width Lis:

$$\begin{gathered} E_n=\frac{P_n^2}{2m} \\ =\frac{n^2{h}^2}{8m{L}^2} \\ =\frac{n^2{\mathrm{\pi }}^2{h}^2}{2m{L}^2}\left(n=1,2,3,...\right) \end{gathered}$$

Determine the width of box.

Given that, energy of photon at ground level is 3.0 eV.

The energy $E_1$in joule is

$$\begin{gathered} E_1=3.00\times \left(\frac{1.60\times {10}^{-19}}{1}\right) \\ =4.80\times {10}^{-19}J \end{gathered}$$

The difference in energy between levels n = 2 and n = 1 is:

$$\begin{gathered} ∆E=E_2-E_1 \\ =\left(2^2-1^2\right)\frac{{\mathrm{\pi }}^2{\mathrm{h}}^2}{2m{L}^2} \\ =3E_1 \\ =\frac{3{\mathrm{\pi }}^2{\mathrm{h}}^2}{2m{L}^2} \end{gathered}$$

The mass of electron is$9.11\times {10}^{-31}kg$and value of$h=1.055\times {10}^{-34}J\cdot s$.

The width of the box is:

$$\begin{gathered} L=\frac{\sqrt{3}n\mathrm{\pi h}}{\sqrt{2}m{E}_n} \\ =\frac{\sqrt{3}\mathrm{\pi }\left(1.055\times {10}^{-34}\right)}{\sqrt{2\left(9.11\times {10}^{-31}\right)\left(4.80\times {10}^{-19}\right)}} \\ =6.14\times {10}^{-10}m \end{gathered}$$

Hence, the width L of the box is $6.14\times {10}^{-10}m$.