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The contact potential between two metal is given by-

${\mathit{V}}_{\mathbf{1}}\mathbf{鈭�}{\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{蠒}}_{\mathbf{2}}\mathbf{鈭�}{\mathbf{蠒}}_{\mathbf{1}}}{\mathbf{e}}$ 鈥�... (1)

Where;

${蠒}_1,{蠒}_2鈫�$Work functions of metal 1 and 2

${\mathit{V}}_{\mathbf{1}}\mathbf{,}{\mathit{V}}_{\mathbf{2}}\mathbf{鈫�}$Potential of metal 1 and 2

$\mathit{e}\mathbf{鈫�}$Elementary unit charge

Also, kineticenergy 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�... (2)

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

Convert given wavelength of metals into standard units

$\begin{array}{l}(410\text{nm})\left(\frac{1脳{10}^{鈭�9}\text{m}}{1\text{nm}}\right)=4.1脳{10}^{鈭�7}\text{m}n\\ (280\text{nm})\left(\frac{1脳{10}^{鈭�9}\text{m}}{1\text{nm}}\right)=2.8脳{10}^{鈭�7}\text{m}\end{array}$

$位$To find the work functions of the metals, equation (2) is used, 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 $蠁$:

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

Substitute $6.63脳{10}^{鈭�34}\text{J}$.s for$h,3脳{10}^8\text{m}/\text{s}$ for$c$, and $4.1脳{10}^7\text{m}$for ,

$\begin{array}{c}{蠁}_1=\frac{\mathrm{鈩�}c}{{\mathrm{鈩�}}_{\max }}\\ =\frac{(6.63脳{10}^{鈭�34}\text{J}.\text{s})(3脳{10}^8\text{m}/\text{s})}{(4.1脳{10}^{鈭�7}\text{m})}\\ =4.85脳{10}^{鈭�19}\text{J}\end{array}$

Substitute $6.63脳{10}^{鈭�34}\text{J}$.s for$h,3脳{10}^8\text{m}/\text{s}$ for$c$, and $2.8脳{10}^{鈭�7}\text{m}$for$位$ ,

$\begin{array}{c}{蠁}_1=\frac{\mathrm{鈩�}c}{{\mathrm{鈩�}}_{\max }}\\ =\frac{(6.63脳{10}^{鈭�34}\text{J}.\text{s})(3脳{10}^8\text{m}/\text{s})}{(2.8脳{10}^{鈭�7}\text{m})}\\ =7.1脳{10}^{鈭�19}\text{J}\end{array}$

Then substitute those into equation (1), along with $1.6脳{10}^{鈭�19}\text{C}$for e:

$\begin{array}{c}{V}_1鈭�V_2=\frac{{蠁}_2鈭�{蠁}_1}{e}\\ =\frac{(7.1脳{10}^{鈭�19}\text{J})(4.85脳{10}^{鈭�19}\text{J})}{(1.6脳{10}^{鈭�19}\text{C})}\\ =1.41\text{V}\end{array}$

Thus, the contact potential between the two metals will be$V_1鈭�V_2=1.4\text{V}$