Question

The photoelectric work function of potassium is 2.3eV. If light that has a wavelength 190nm of falls on potassium, find (a) the stopping potential in volts; (b) the kinetic energy, in electron volts, of the most energetic electrons ejected; (c) the speed of these electrons.

Short answer

1. The stopping potential is $\mathbf{4}\mathbf{.}\mathbf{24}\mathit{e}\mathit{V}$.
2. The kinetic energy of the most energetic electrons ejected is $\mathbf{4}\mathbf{.}\mathbf{24}\mathit{e}\mathit{V}$.
3. The speed of theses electrons is $\mathbf{1}\mathbf{.}\mathbf{22}\mathbf{*}{\mathbf{10}}^{\mathbf{6}}\mathit{m}\mathbf{/}\mathit{s}$.

Step-by-step solution

Formula for maximum kinetic energy

${\mathit{K}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{2}\mathbf{=}\mathit{h}\mathit{f}\mathbf{-}\mathit{\varphi }$ (1)

$\mathrm{where}{K}_{max}=e{V}_0$

$\mathbf{\Rightarrow }\mathit{e}{\mathit{V}}_{\mathbf{0}}\mathbf{=}\mathit{h}\mathit{f}\mathbf{-}\mathit{\varphi }\mathbf{=}\mathit{h}\frac{\mathbf{c}}{\mathbf{\lambda }}\mathbf{-}\mathit{\varphi }$

Calculate the stopping potential

Given: $\mathit{e}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{C}$

$$\begin{gathered} \mathit{\lambda }\mathbf{=}\mathbf{190}\mathit{n}\mathit{m}\mathbf{=}\mathbf{190}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{C} \\ \mathit{h}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{34}}\mathit{J}\mathit{s} \\ \mathit{\varphi }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{3}\mathit{e}\mathit{V} \end{gathered}$$

Substitute the given in equation (2),

$$\begin{gathered} {\mathit{V}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{e}}\mathbf{\left(}\frac{\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{34}}\mathbf{J}\mathbf{s}\mathbf{*}\mathbf{3}\mathbf{*}{\mathbf{10}}^{\mathbf{8}}\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{190}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{m}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{54}\mathit{e}\mathit{V}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{3}\mathit{e}\mathit{V} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{24}\mathit{e}\mathit{V} \end{gathered}$$

Calculate the kinetic energy

Substitute the values in equation (1),

$$\begin{gathered} {\mathit{K}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\frac{\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{34}}\mathbf{J}\mathbf{s}\mathbf{*}\mathbf{3}\mathbf{*}{\mathbf{10}}^{\mathbf{8}}\mathbf{J}\mathbf{s}}{\mathbf{190}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{m}}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{3}\mathit{e}\mathit{V} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{54}\mathit{e}\mathit{V}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{3}\mathit{e}\mathit{V} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{24}\mathit{e}\mathit{V} \end{gathered}$$

Calculate the speed of electrons

From equation (1), the speed of electrons can be given as:

$$\begin{gathered} \mathit{v}\mathbf{=}\sqrt{\frac{\mathbf{2}}{\mathbf{m}}\mathbf{(}\frac{\mathbf{h}\mathbf{c}}{\mathbf{\lambda }}\mathbf{-}\mathbf{\varphi }\mathbf{)}} \\ \mathbf{}\mathbf{}\mathbf{=}\sqrt{\frac{\mathbf{2}}{\mathbf{9}\mathbf{.}\mathbf{1}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{31}}\mathbf{k}\mathbf{g}}\mathbf{(}\frac{\mathbf{6}\mathbf{.}\mathbf{26}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{34}}\mathbf{J}\mathbf{s}\mathbf{*}\mathbf{3}\mathbf{*}{\mathbf{10}}^{\mathbf{8}}\mathbf{J}\mathbf{s}}{\mathbf{190}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{m}}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{e}\mathbf{V}\mathbf{)}} \\ \mathbf{}\mathbf{}\mathbf{=}\sqrt{\frac{\mathbf{2}}{\mathbf{9}\mathbf{.}\mathbf{1}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{31}}\mathbf{k}\mathbf{g}}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{78}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{J}\mathbf{)}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{22}\mathbf{*}{\mathbf{10}}^{\mathbf{6}}\mathit{m}\mathbf{/}\mathit{s} \end{gathered}$$

Thus, the stopping potential is 4.24 eV. The kinetic energy of the most energetic electrons ejected is 4.24 eV. The speed of theses electrons is 1.22*${\mathbf{10}}^{\mathbf{6}}\mathit{m}\mathbf{/}\mathit{s}$.