The smallest amount of energy needed to eject an electron from metallic sodium is $\mathbf{2}\mathbf{.}\mathbf{28}\mathbf{eV}$. Does sodium show a photoelectric effect for red light, with $\lambda \mathbf{=}\mathbf{680}\mathbf{nm}$? (That is, does the light cause electron emission?) (b) What is the cutoff wavelength for photoelectric emission from sodium? (c) To what color does that wavelength correspond?

The given data can be listed below as,

• Thesmallest amount of energy needed to eject an electron from metallic sodium is$A=2.28\text{eV}$.
• The wavelength of red light is, $\lambda =680\text{nm}$.

In this question, the value of the light wavelength of an electron can be obtained with the help of the value of the electron’s energy and Plank’s constant and light speed in a vacuum. The relation between the light energy and its wavelength is an indirect linear one.

The expression to calculate the energy carried by the photon with wavelength 680nm is expressed as,

$E=\frac{hc}{\lambda }$

Here, E is the energy carried by the photon with wavelength 680 nm , h is the Plank’s constant whose value is $6.63\times {10}^{-34}\text{J}·\text{s}$, c is the light speed in vacuum whose value is $3\times {10}^8\text{m/s}$.

Substitute all the known values in the above equation.

$\begin{array}{rcl}E& =& \left[\frac{\left(6.63\times {10}^{-34}\text{J}·\text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(680\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\right]\\ & \approx & 2.93\times {10}^{-19}\text{J}\\ & \approx & \left(2.93\times {10}^{-19}\text{J}\right)\left(\frac{\frac{1}{1.6\times {10}^{-19}}\text{eV}}{1\text{J}}\right)\\ & \approx & 1.83\text{eV}\end{array}$

From the above calculations, the energy carried by the photon with wavelength 680 nm is less than the smallest amount of energy needed to eject an electron from metallic sodium $\left(2.28\text{eV}\right)$(work function). Therefore, there is no photoelectric emission takes place.

Thus, the sodium does not show a photoelectric effect for the red light.

The expression to calculate the cut-off wavelength for photoelectric emission from sodium is expressed as,

$$\begin{gathered} \frac{hc}{{\lambda }_o}=A \\ {\lambda }_o=\frac{hc}{A} \end{gathered}$$

Here, ${\lambda }_o$is the cut-off wavelength for photoelectric emission from sodium.

Substitute all the known values in the above equation.

$\begin{array}{rcl}{\lambda }_o& =& \left[\frac{\left(6.63\times {10}^{-34}\text{J}·\text{s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(2.28\text{eV}\right)\left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right)}\right]\\ & \approx & 5.45\times {10}^{-7}\text{m}\\ & \approx & \left(5.45\times {10}^{-7}\text{m}\right)\times \left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ & \approx & 545\text{nm}\end{array}$

Thus, the cut-off wavelength for photoelectric emission from sodium is 545 nm.

From the above calculations, one can observe that the cut-off wavelength of the photoelectric effect from sodium is less than the wavelength of the red light. The value of the cut-off wavelength of the photoelectric effect from sodium comes in the range of the wavelength of the green color.

Thus, the color of the corresponding wavelength is green.