Q12.122CP - Chemistry: Molecular Nature Of Matter And Change

Question

In the photoelectric effect, the work function (Φ) is the minimum energy a photon must have to remove an electron from a metal surface (see Problem 7.70). For a given metal, Φ depends on how and where the photon strikes the crystal. Copper adopts the face-centered cubic structure. If the photon strikes perpendicular to the unit cell’s face, Φ is 4.59 eV; if perpendicular to a plane tangent to an edge, Φ is 4.48 eV; and if perpendicular to a plane tangent to a corner, Φ is 4.94 eV (1 eV =  $\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{-19}\mathbf{J}$). Find the wavelength (in nm) of the lowest energy photon that can remove an electron from copper, and state how it strikes the unit cell.

Step-by-Step Solution

Answer

The wavelength decreases as the work function increases.

 

 $\begin{array}{l}{\text{ν}}_{\text{0}}=4.59\text{eV}\\ \text{λ}=2.7\times {10}^{-7}\text{m}\end{array}$

 

 $\begin{array}{c}{\text{ν}}_{\text{0}}=4.48\text{eV}\\ \text{λ}=2.8\times {10}^{-7}\text{m}\end{array}$

 

 $\begin{array}{c}{\text{ν}}_{\text{0}}=4.94\text{eV}\\ \text{λ}=2.51\times {10}^{-7}\text{m}\end{array}$

 

The wavelength (in nm) of the lowest energy photon with the energy of  $7.168\times {10}^{-19}\mathrm{J}$ that can remove an electron from copper is 0.028nm.

It strikes the unit cell with 

Step 1: Definition

The photoelectric work function is the minimum energy that is required in ejecting an electron from the surface of a metal.

The striking of the lowest energy photon that can remove an electron from a copper metal surface is called the photoelectric effect. The striking takes place with the energy,

 $\text{E = hνo}$

Work function is a form of energy required for a wave to propagate in space. It can also be defined as the multiplication of the plank’s constant and frequency.

 

Frequency can be defined as the periodic repetition of the wave at a particular set of times. The frequency can be defined as the speed of the light to the wavelength of the wave.

$\mathbf{\nu o}\mathbf{=}\frac{c}{\lambda }$

$\text{E = h}\frac{c}{\lambda }$

Also,

Plank’s constant,  $\mathrm{h}=6.6\times {10}^{-34}\mathrm{Js}$

Speed of light $=3\times {10}^8{\mathrm{ms}}^{-1}$ 

Step 2: The wavelength, when the work function of the wave is 4.59eV.

The work function from eV to Joule.

 

 $\begin{array}{c}1\mathrm{eV}=1.602\times {10}^{-19}\mathrm{J}\\ 4.59\mathrm{eV}=4.59\mathrm{eV}\times \frac{1.602\times {10}^{-19}\mathrm{J}}{1\mathrm{eV}}\\ 4.59\mathrm{eV}=7.35\times {10}^{-19}\mathrm{J}\end{array}$

 

Also,

                        $\text{E}\text{ = h}\frac{c}{\lambda }$

$\begin{array}{c}7.35\times {10}^{-19}\mathrm{J}=\frac{6.6\times {10}^{34}\mathrm{Js}\times 3\times {10}^8{\mathrm{ms}}^{-1}}{\mathrm{\lambda }}\\ \mathrm{\lambda }=\frac{6.6\times {10}^{34}\mathrm{Js}\times 3\times {10}^8{\mathrm{ms}}^{-1}}{7.35\times {10}^{-19}\mathrm{J}}\\ \mathrm{\lambda }=2.7\times {10}^{-7}\mathrm{m}\end{array}$

 

 

Therefore, the wave length when the work function is $4.59\mathrm{eV}$  is  $2.7\times {10}^{-7}\mathrm{m}$.

Step 3: The wavelength, when the work function of the wave is 4.48eV.

The work function from eV to Joule.

 

 $\begin{array}{c}1\mathrm{eV}=1.602\times {10}^{-19}\mathrm{J}\\ 4.48\mathrm{eV}=4.48\mathrm{eV}\times \frac{1.602\times {10}^{-19}\mathrm{J}}{1\mathrm{eV}}\\ 4.48\mathrm{eV}=7.168\times {10}^{-19}\mathrm{J}\end{array}$

 

Also,

                          $\text{E}\text{ = h}\frac{c}{\lambda }$

$\begin{array}{c}7.168\times {10}^{-19}\mathrm{J}=\frac{6.6\times {10}^{34}\mathrm{Js}\times 3\times {10}^8{\mathrm{ms}}^{-1}}{\mathrm{\lambda }}\\ \mathrm{\lambda }=\frac{6.6\times {10}^{34}\mathrm{Js}\times 3\times {10}^8{\mathrm{ms}}^{-1}}{7.168\times {10}^{-19}\mathrm{J}}\\ \mathrm{\lambda }=2.8\times {10}^{-7}\mathrm{m}\end{array}$

Therefore, the wave length when the work function is 4.48eV is  ${\text{2.8 × 10}}^{\text{-7}}\text{m}$.

 

Step 4: The wavelength, when the work function of the wave is 4.94eV.

The work function from eV to Joule.

 

 $\begin{array}{c}{\text{1eV = 1.602 × 10}}^{\text{-19}}\text{J}\\ \text{4.94eV =4.94eV × }\frac{{\text{1.602×10}}^{\text{-19}}\text{J}}{\text{1eV}}\\ {\text{4.94eV =7.9 × 10}}^{\text{-19}}\text{J}\end{array}$

 

Also,

 

                    $\text{E}\text{ = h}\frac{c}{\lambda }$

 $\begin{array}{c}7.9\times {10}^{-19}\mathrm{J}=\frac{6.6\times {10}^{34}\mathrm{Js}\times 3\times {10}^8{\mathrm{ms}}^{-1}}{\mathrm{\lambda }}\\ \mathrm{\lambda }=\frac{6.6\times {10}^{34}\mathrm{Js}\times 3\times {10}^8{\mathrm{ms}}^{-1}}{7.9\times {10}^{-19}\mathrm{J}}\\ \mathrm{\lambda }=2.51\times {10}^{-7}\mathrm{m}\end{array}$

 

Therefore, the wave length when the work function is 4.49eV is ${\text{2.51×10}}^{\text{-7}}\text{m}$ .

 

The lowest energy photon that requires to remove an electron from copper is $7.168\times {10}^{-19}\mathrm{J}$   having work function  $4.48\mathrm{eV}=7.168\times {10}^{-19}\mathrm{J}$ from the plane tangent to an edge and the wavelength is  ${\text{2.8 × 10}}^{\text{-7}}\text{m = 0.028nm}$ .

 

The striking of the lowest energy photon that can remove an electron from a copper metal surface is called the photoelectric effect. The striking takes place with the energy,

 $\text{E = hνo}$