With light of wavelength $\mathbf{520}\mathbf{}\mathbf{\text{nm}}$ . Photoelectrons are ejected from a metal surface with a maximum speed of $\mathbf{1}\mathbf{.78}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{\text{m}}\mathbf{/}\mathbf{\text{s}}$.

(a) What wavelength would be needed to give a maximum speed of $\mathbf{4}\mathbf{.81}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{\text{m}}\mathbf{/}\mathbf{\text{s}}$?

(b) Can you guess what metal it is?

$\begin{array}{c}\text{Speed\hspace{0.17em}of\hspace{0.17em}photoelectrons\hspace{0.17em}}{\mathrm{v}}_1=1.78\times {10}^5\frac{\text{m}}{\text{s}}\text{\hspace{0.17em}\hspace{0.17em}at}{\mathrm{\lambda }}_1=520\text{\hspace{0.17em}nm}\\ \text{Speed\hspace{0.17em}required\hspace{0.17em}}{\mathrm{v}}_2=4.81\times {10}^5\frac{\text{m}}{\text{s}}\end{array}$

$\text{Planck\hspace{0.17em}constant\hspace{0.17em}}\mathrm{h}=1240\text{\hspace{0.17em}ev}\cdot \text{nm}$

$\text{Energy\hspace{0.17em}released\hspace{0.17em}}\mathrm{c}=1.6\times {10}^{-19}\frac{\text{J}}{\text{eV}}$

Determine the formula for the kinetic energy and the work done as:

$\begin{array}{l}{\mathbf{KE}}_{\mathbf{max}\mathbf{1}}\mathbf{=}{\mathbf{hf}}_{\mathbf{1}}\mathbf{-}\mathbf{\varphi }\\ \frac{{\mathbf{m}}_{\mathbf{e}}{\mathbf{v}}_{\mathbf{1}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{=}\frac{\mathbf{hc}}{{\mathbf{\lambda }}_{\mathbf{1}}}\mathbf{-}\mathbf{\varphi }\end{array}$

a)

However, the two tests are carried out using the same metal, therefore the work function will be the same in both situations. Here, there are two independent experiments; the first one is carried out using a light of wavelength, while the other wavelength is unknown.

$\begin{array}{l}{\mathrm{KE}}_{\max 1}={\mathrm{hf}}_1-\mathrm{\varphi }\\ \frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}_1^2}{2}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_1}-\mathrm{\varphi }\\ {\mathrm{KE}}_{\max 2}={\mathrm{hf}}_2-\mathrm{\varphi }\\ \frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}_2^2}{2}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_2}-\mathrm{\varphi }\end{array}$

Solve further as:

$\begin{array}{c}\frac{{\mathrm{m}}_{\mathrm{e}}}{2}({\mathrm{v}}_2^2-{\mathrm{v}}_1^2)=\mathrm{hc}\left(\frac{1}{{\mathrm{\lambda }}_2}-\frac{1}{{\mathrm{\lambda }}_1}\right)\\ \left(\frac{1}{{\mathrm{\lambda }}_2}-\frac{1}{{\mathrm{\lambda }}_1}\right)=\frac{\frac{{\mathrm{m}}_{\mathrm{e}}}{2}({\mathrm{v}}_2^2-{\mathrm{v}}_1^2)}{\mathrm{hc}}\\ \frac{1}{{\mathrm{\lambda }}_2}=\frac{\frac{{\mathrm{m}}_{\mathrm{e}}}{2}({\mathrm{v}}_2^2-{\mathrm{v}}_1^2)}{\mathrm{hc}}+\frac{1}{{\mathrm{\lambda }}_1}\end{array}$

Substitute the values

$\begin{array}{c}\frac{1}{{\lambda }_2}=\frac{\frac{9.1\times {10}^{-31}\text{kg}}{2}\times \left({\left(4.81\times {10}^5\frac{\text{m}}{\text{s}}\right)}^2-{\left(1.78\times {10}^5\frac{\text{m}}{\text{s}}\right)}^2\right)}{1240\text{eV}.\text{nm}\times 1.6\times {10}^{-19}\frac{\text{J}}{\text{eV}}}+\frac{1}{520\text{nm}}\frac{9.08\times {10}^{-20}\text{J}}{1}+\frac{1}{520\text{nm}}\\ \frac{1}{{\lambda }_2}\approx \frac{1}{0.00238}\frac{1}{\text{nm}}\\ \frac{1}{{\lambda }_2}=\frac{1}{1.984\times {10}^{-16}\text{J}.\text{nm}}\\ {\lambda }_2=420\text{nm}\end{array}$

b)

The identical findings would be obtained by first calculating the work function and then determining the wavelength. However, locate the metal that corresponds to this number by solving for the work function of this metal using equation (1) or (2), then comparing the results with the table.

$\begin{array}{l}{\mathrm{KE}}_{\max 1}={\mathrm{hf}}_1-\mathrm{\varphi }\\ \frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}_1^2}{2}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_1}-\mathrm{\varphi }\\ \mathrm{\varphi }=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_1}-\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}_1^2}{2}\\ \mathrm{\varphi }=\frac{1240\text{\hspace{0.17em}eV}\cdot \text{nm}}{520\text{nm}}-\frac{9.1\times {10}^{-31}\text{kg}\times {\left(1.78\times {10}^5\frac{\text{m}}{\text{s}}\right)}^2}{2\times 1.6\times {10}^{-19}\frac{\text{J}}{\text{eV}}}\end{array}$

Solve further as:

$\begin{array}{l}\mathrm{\varphi }=2.384\text{eV}-0.090\text{eV}\\ \mathrm{\varphi }\approx \mathbf{2}.\mathbf{3}\text{\hspace{0.17em}eV}\end{array}$

From table $3.1$, this is exactly the same work function of Sodium.

Hence, this metal is most probably Sodium.