91Ó°ÊÓ

${\mathit{K}}_{\mathbf{max}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}_{\mathbf{max}}\mathbf{2}\mathbf{=}\mathit{h}\mathit{f}\mathbf{-}\mathit{Ï•}$ (1)

where ${\mathit{K}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\mathit{e}{\mathit{V}}_{\mathbf{0}}$

$\mathbf{⇒}\mathit{e}{\mathit{V}}_{\mathbf{0}}\mathbf{=}\mathit{h}\mathit{f}\mathbf{-}\mathit{ϕ}\mathbf{=}\mathit{h}\frac{\mathbf{c}}{\mathbf{λ}}\mathbf{-}\mathit{ϕ}$

Calculate${\mathit{λ}}_{\mathbf{t}\mathbf{h}}$ using the following equation,

${\mathit{λ}}_{\mathbf{th}}\mathbf{=}\frac{\mathbf{c}}{{\mathbf{f}}_{\mathbf{t}\mathbf{h}}}$ (3)

To find ${\mathit{λ}}_{\mathbf{th}}$,use the equation given below,

${\mathit{K}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\mathit{h}{\mathit{f}}_{\mathbf{t}\mathbf{h}}\mathbf{-}\mathit{Ï•}$

At threshold, ${\mathit{K}}_{\mathbf{max}}\mathbf{=}\mathbf{0}$

Thus,

$\mathit{h}{\mathit{f}}_{\mathbf{th}}\mathbf{-}\mathit{Ï•}$ (4)

Now, using equation (2),

$\mathbf{⇒}\mathit{ϕ}\mathbf{=}\left(6.626*{10}^{-34}*1.18*{10}^{15}\right)\mathbf{-}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{*}\mathbf{0}\mathbf{.}\mathbf{18}\mathbf{)}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{53}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{J}$

Now, $\mathbf{1}\mathit{e}\mathit{V}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{J}$

$\mathbf{⇒}\mathit{ϕ}\mathbf{=}\frac{\mathbf{7}\mathbf{.}\mathbf{53}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}}{\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{7}\mathit{e}\mathit{V}$

Substitute the value in equation (4),

${\mathit{f}}_{\mathbf{t}\mathbf{h}}\mathbf{=}\frac{\mathbf{7}\mathbf{.}\mathbf{53}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}}{\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{34}}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{14}\mathbf{*}{\mathbf{10}}^{\mathbf{15}}\mathit{H}\mathit{z}$

Substitute the value in equation (3),

$$\begin{gathered} {\mathit{λ}}_{\mathbf{th}}\mathbf{=}\frac{\mathbf{3}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}}{\mathbf{1}\mathbf{.}\mathbf{14}\mathbf{*}{\mathbf{10}}^{\mathbf{15}}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{64}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathit{m} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{264}\mathit{n}\mathit{m} \end{gathered}$$

Thus, the photoelectric threshold wavelength for the copper surface is 264nm . The work function for this surface is 4.7eV . The calculated value agrees with the value in the table.