The maximum wavelength light that will eject electrons from metal I via the photoelectric effect is$\mathbf{410}\mathbf{}\mathbf{\text{nm}}$. For metal2, it is$\mathbf{280}\mathbf{}\mathbf{\text{nm}}$. What would be the potential difference if these two metals were put in contact?

The contact potential between two metal is given by-

${\mathit{V}}_{\mathbf{1}}\mathbf{-}{\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{\varphi }}_{\mathbf{2}}\mathbf{-}{\mathbf{\varphi }}_{\mathbf{1}}}{\mathbf{e}}$ …... (1)

Where;

${\varphi }_1,{\varphi }_2\to$Work functions of metal 1 and 2

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

$\mathit{e}\mathbf{\to }$Elementary unit charge

Also, kineticenergy of photons ejected while being struck by photons is-

$\mathit{K}\mathit{E}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{\lambda }}\mathbf{-}\mathit{\varphi }$ …... (2)

Where;

$\mathit{h}\mathbf{\to }$Planck's constant

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

$\mathit{\varphi }\mathbf{\to }$Work function of material

$\mathit{\lambda }\mathbf{\to }$Wavelength of photon

Convert given wavelength of metals into standard units

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

$\lambda$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${\lambda }_{\max }$then solved for $\phi$:

$\begin{array}{c}KE=\frac{hc}{\lambda }-\phi 0\\ =\frac{hc}{{\lambda }_{\max }}-\phi \phi \\ =\frac{hc}{{\lambda }_{\max }}\end{array}$

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

$\begin{array}{c}{\phi }_1=\frac{\hslash c}{{\hslash }_{\max }}\\ =\frac{(6.63\times {10}^{-34}\text{J}.\text{s})(3\times {10}^8\text{m}/\text{s})}{(4.1\times {10}^{-7}\text{m})}\\ =4.85\times {10}^{-19}\text{J}\end{array}$

Substitute $6.63\times {10}^{-34}\text{J}$.s for$h,3\times {10}^8\text{m}/\text{s}$ for$c$, and $2.8\times {10}^{-7}\text{m}$for$\lambda$ ,

$\begin{array}{c}{\phi }_1=\frac{\hslash c}{{\hslash }_{\max }}\\ =\frac{(6.63\times {10}^{-34}\text{J}.\text{s})(3\times {10}^8\text{m}/\text{s})}{(2.8\times {10}^{-7}\text{m})}\\ =7.1\times {10}^{-19}\text{J}\end{array}$

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

$\begin{array}{c}{V}_1-V_2=\frac{{\phi }_2-{\phi }_1}{e}\\ =\frac{(7.1\times {10}^{-19}\text{J})(4.85\times {10}^{-19}\text{J})}{(1.6\times {10}^{-19}\text{C})}\\ =1.41\text{V}\end{array}$

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