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The total depth of a potential well${\mathit{U}}_{\mathbf{0}}$for a material just the sum of its Fermi energy${\mathit{E}}_{\mathbf{F}}$and its work function$\mathit{蠒}$

${\mathit{U}}_{\mathbf{0}}\mathbf{=}{\mathit{E}}_{\mathbf{F}}\mathbf{+}\mathit{蠒}\mathbf{\text{鈥�}}$

鈥�.. (1)

The Fermi energy${\mathit{E}}_{\mathbf{F}}$of a materialis given by:

${\mathit{E}}_{\mathbf{F}}\mathbf{=}\frac{{\mathbf{蟺}}^{\mathbf{2}}{\mathbf{鈩�}}^{\mathbf{2}}}{\mathbf{m}}{\mathbf{\left[}\frac{\mathbf{3}}{\mathbf{(}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{蟺}\sqrt{\mathbf{2}}\mathbf{V}}\frac{\mathbf{N}}{\mathbf{V}}\mathbf{\right]}}^{\mathbf{2}\mathbf{/}\mathbf{3}}$ 鈥�.. (2)

Also, kinetic energy of photons ejected while being struck by photons is-

$\mathit{K}\mathit{E}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{位}}\mathbf{鈭�}\mathit{蠒}$ 鈥�.. (3)

Where;

$\mathit{h}\mathbf{鈫�}$Planck's constant

$\mathit{c}\mathbf{鈫�}$Speed of light in vacuum

$\mathit{蠒}\mathbf{鈫�}$Work function of material

$\mathit{位}\mathbf{鈫�}$Wavelength of photon

Given Information:

The density of copper$=8.9脳{10}^3\text{kg}/{\text{m}}^3$

The wavelength of photo$=275\text{nm}$

Calculation:

role="math" localid="1658391895008" $\begin{array}{c}{E}_F=\frac{{蟺}^2{\mathrm{鈩�}}^2}{m}{\left[\frac{3}{(2s+1)蟺\sqrt{2}V}\frac{N}{V}\right]}^{\frac{2}{3}}\\ =\frac{{蟺}^2{\mathrm{鈩�}}^2}{m}{\left[\frac{3}{(2s+1)蟺\sqrt{2}V}\frac{D}{m位}\right]}^{\frac{2}{3}}\end{array}$

Substitute$9.1脳{10}^{鈭�31}\text{kg}$ for$\text{m},1.055脳{10}^{鈭�34}\text{J}.\text{s}$ for $\text{h};\frac{1}{2}$for$\text{s},8.9脳{10}^3$

$\text{kg}/{\text{m}}^3$for $D$, $1.055脳{10}^{鈭�25}\text{kg}$and for$m_{鈭�}$ ,

$\begin{array}{c}{E}_F=\frac{{蟺}^2{\mathrm{鈩�}}^2}{m}{\left[\frac{3}{(2s+1)蟺\sqrt{2}}\frac{D}{m位}\right]}^{\frac{2}{3}}\\ =\frac{{蟺}^2{(1.055脳{10}^{鈭�34}\text{J}.\text{s})}^2}{(9.1脳{10}^{鈭�31}\text{kg})}{\left[\frac{3}{(2\left[\frac{1}{2}\right]+1)蟺\sqrt{2}}\frac{(8.9脳{10}^3\frac{\text{kg}}{{\text{m}}^3})}{(1.055脳{10}^{鈭�25}\text{kg})}\right]}^{\frac{2}{3}}\\ =1.126脳{10}^{鈭�18}\text{J}\end{array}$

Then the work function needs to be found by the use of equation (3), with the condition that the kinetic energy of the ejected electrons will be 0 for the maximum possible wavelength used${位}_{\max }\$,$then solved for $蠒$:

$\begin{array}{c}KE=\frac{hc}{位}鈭�蠒0\\ =\frac{hc}{{位}_{\max }}鈭�蠒蠁\\ =\frac{hc}{{位}_{\max }}\end{array}$

Substitute $2.75脳{10}^{鈭�7}\text{m}$for${位}_{\max },6.63脳{10}^{鈭�34}\text{J}.\text{s}$ , and $3.0脳{10}^8\text{m}/\text{s}$for c,

$\begin{array}{c}蠁=\frac{hc}{{位}_{\max }}\\ =\frac{(6.63脳{10}^{鈭�34}\text{J}.\text{s})(3脳{10}^8\frac{\text{m}}{\text{s}})}{(2.75脳{10}^{鈭�7}\text{m})}\\ =7.233脳{10}^{鈭�19}\text{J}\end{array}$

Now the Fermi energy $(1.126脳{10}^{鈭�18}\text{J})$and work function$(7.233脳{10}^{鈭�19}\text{J}7.233*{10}^{鈭�19}\text{J})$ can be combined in equation (1) to get the potential:

$\begin{array}{c}{U}_0=E_F+蠁\\ =(1.126脳{10}^{鈭�18}\text{J})+(7.233脳{10}^{鈭�19}\text{J})\\ =1.85脳{10}^{鈭�18}\text{J}\end{array}$

That can be converted to :

$\begin{array}{c}{U}_0=(1.85脳{10}^{鈭�18}\text{J})\left(\frac{1\text{eV}}{1.6脳{10}^{鈭�19}\text{J}}\right)\\ =11.55\text{eV}\end{array}$