Question

(a) Find the excitation energy from the ground level to the third excited level for an electron confined to a box of width$\mathbf{0}\mathbf{.}\mathbf{360}\mathbf{}\mathbf{nm}$. (b) The electron makes a transition from the$\mathit{n}\mathbf{=}\mathbf{1}$to$\mathit{n}\mathbf{=}\mathbf{4}$level by absorbing a photon. Calculate the wavelength of this photon.

Short answer

a) The excitation energy from the ground level $n=1$to the third excited level $n=4$is $6.97\times {10}^{-18}\mathrm{J}$

b) The wavelength of the photon is $28.5\mathrm{nm}$.

Step-by-step solution

Define the energy level.

The energy level of a particle of mass$\mathit{m}$in a box of width$\mathit{L}$is:

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

The energy of photon$E=\frac{hc}{\lambda }$where,$\mathbf{\lambda }$is the wavelength.

Determine the excitation energy.

Given that, the width of box is$0.360\mathrm{nm}=0.360\times {10}^{-9}\mathrm{m}$.

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

The excitation energy from the ground level $n=1$to the third excited level$n=2$:

localid="1663994790127" $$\begin{gathered} ∆E=E_4-E_1 \\ =\left(4^2-4^1\right)\frac{{\pi }^2{h}^2}{2mL} \\ =15\frac{{\pi }^2{h}^2}{2mL} \\ =15\frac{{\mathrm{\pi }}^2\left(1.055\times {10}^{-34}\right)}{2\left(9.11\times {10}^{-31}\right){\left(0.360\times {10}^{-9}\right)}^2} \end{gathered}$$

Hence, excitation energy from the ground level $n=1$to the third excited level $n=4$isrole="math" localid="1663995018936" $6.97\times {10}^{-18}\mathrm{J}$

Determine the wavelength.

The speed $c$of light is $3.00\times {10}^8$$\mathrm{m}/\mathrm{s}$and the Planck’s constant is$6.97\times {10}^{-18}\mathrm{J}.\mathrm{s}$

The wavelength of photon$\lambda$:

$$\begin{gathered} \lambda =\frac{hc}{∆E} \\ =\frac{\left(6.626\times {10}^{-34}\right)\left(3\times {10}^8\right)}{6.97\times {10}^{-18}} \\ =2.85\times {10}^{-8}\mathrm{m} \\ =2.85\mathrm{nm} \end{gathered}$$

Hence, wavelength of the photon is$2.85\mathrm{nm}$.