Question

An electron, trapped in a one-dimensional infinite potential well 250 pm wide, is in its ground state. How much energy must it absorb if it is to jump up to the state with $\mathbf{n}\mathbf{=}\mathbf{4}$?

Short answer

90.3 eV

Step-by-step solution

Given data

The width of the potential well is

$$\begin{gathered} L=250\mathrm{pm} \\ =0.250\mathrm{nm} \\ n=4 \end{gathered}$$

Energy in a potential well

The $n^{th}$ state energy of a particle of mass in an infinite potential well of width L is

$E_n=\left(\frac{n^2{h}^2}{8mL}\right)..........\left(1\right)$

Determining the energy required to jump from ground state to the fourth state

Here h is the Planck's constant having value

$$\begin{gathered} h=6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}\mathrm{and}\mathrm{c}=2.998\times {10}^8\mathrm{m}/\mathrm{s} \\ {\mathrm{h}}_{\mathrm{c}}=\frac{\left(6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(2.998\times {10}^8\mathrm{m}/\mathrm{s}\right)}{\left(1.602\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)\left({10}^{-9}\mathrm{m}/\mathrm{nm}\right)} \\ =1240\mathrm{eV}.\mathrm{nm} \end{gathered}$$

Using the value for an electron from Table 37-3 $\left(511\times {10}^3\mathrm{eV}\right)$, Eq.39 – 4 can be rewritten as

$$\begin{gathered} E_n=\frac{n^2{h}^2}{8m{L}^2} \\ =\frac{n^2{\left(h_c\right)}^2}{8\left(m{c}^2\right)L^2} \end{gathered}$$

The absorbed energy is

$$\begin{gathered} ∆E=E_4-E_1 \\ =\frac{\left(4^2-1^2\right){\left({\mathrm{h}}_{\mathrm{c}}\right)}^2}{8\left({\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2\right){\mathrm{L}}^2} \\ =\frac{15{\left(1240\mathrm{eV}.\mathrm{nm}\right)}^2}{8\left(511\times {10}^3\mathrm{eV}\right){\left(0.250\mathrm{nm}\right)}^2} \\ =90.3\mathrm{eV} \end{gathered}$$

Hence, the required energy is $90.3\mathrm{eV}$