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

$$\begin{gathered} {\mathit{E}}_{\mathbf{n}}\mathbf{=}\frac{{{\mathbf{P}}_{\mathbf{n}}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{n}}^{\mathbf{2}}{\mathbf{h}}^{\mathbf{2}}}{\mathbf{8}\mathbf{m}\mathbf{L}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{n}}^{\mathbf{2}}{\mathbf{Ï€}}^{\mathbf{2}}{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}\mathbf{L}}\left(n=1,2,3...\right) \end{gathered}$$

The ground state energy as a fraction of the ground-level energy for an infinitely deep well $\left(E_{1-1DW}\right)$is given by:

$E_1=0.625E_{1-1DW}$

The difference between the energy of the ground level and the depth of the potential well is equal to the minimum energy required for photon to release the electron.

$E_{min}=U_0-E_1$

The energy of photon $\mathit{E}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{λ}}$where, $\mathbf{λ}$is the wavelength.

Given that, the energy level of square well of depth is$U_0=6E_{1-1DW}$and the width of well is$1.50\mathrm{nm}=1.50×{10}^{-9}\mathrm{m}$.

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

The minimum energy$E_{min}$:

localid="1664000583493" $$\begin{gathered} E_{min}=6E_{1-1\mathrm{DW}}-0.625E_{1-1\mathrm{DW}} \\ =5.375E_{1-1\mathrm{DW}} \\ =\frac{5.375{\mathrm{π}}^2{\mathrm{h}}^2}{2m{L}^2} \\ =\frac{5.375{\mathrm{π}}^2\left(1.055×{10}^{-34}\right)}{2\left(9.11×{10}^{-31}\right)\left(1.50×{10}^{-9}\right)} \\ =1.44×{10}^{-19}\mathrm{J} \end{gathered}$$

Hence, the minimum energy required for photon to release the electron is$1.44×{10}^{-19}\mathrm{J}$.

The speed $c$of light is $3.00×{10}^8\mathrm{m}/\mathrm{s}$and the Planck’s constant is localid="1664000733663" $6.626×{10}^{34}\mathrm{J}â‹\dots \mathrm{s}$.

The wavelength of photon $\mathrm{λ}$is:

$\mathrm{λ}=\frac{hc}{E_{min}}$

$$\begin{gathered} =\frac{\left(6.26×{10}^{-34}\right)\left(3.00×{10}^8\right)}{1.44×{10}^{-19}} \\ =1.38×{10}^6\mathrm{m} \\ =1.38\mathrm{μ}\mathrm{m} \end{gathered}$$

Hence, the maximum wavelength can the photon have and still liberate the electron from the well is $1.38\mathrm{μ}\mathrm{m}$.