Chapter 7, Problem 7.88QE

Interpretation Introduction

Interpretation:

The photoelectric frequency for potassium has to be determined.

Concept Introduction:

The wave nature of any light can be described by its frequency, wavelength, and amplitude. The wavelength $\left(\text{λ, lambda}\right)$ of the light is defined as the distance between two successive peaks. Its SI unit is meter. The frequency $\left(v\right)$ is defined as the number of waves that can pass through the one point in 1 second. Its SI unit is ${\text{s}}^{-1}$. The amplitude $\left(A\right)$ of the light wave is the maximum height of a wave.

The relation between frequency $\left(v\right)$ and wavelength $\left(\text{λ}\right)$ of the light is as follows:

  $v=\frac{c}{\text{λ}}$        (1)

Here, $c$ is the speed of light and its value is $3.00\times {10}^8\text{ m/s}$.

When the electron is ejected from a solid surface because of exposure to the light, this process is known as the photoelectric effect. As the tiny particles of light known as photons having energy $hv$ strike the surface, some of its energy $h{v}_0$ used to overcome the attraction between the electron and solid. The rest of energy gives kinetic energy to the electron. The overall energy expression for photoelectric effect can be shown as follows:

  $hv=h{v}_0+\text{KE}$        (2)

Here $v$ is the frequency of the photon, $v_0$ is the photoelectric threshold frequency, $h$ is known as Plank’s constant and its value is $\text{6}\text{.626}\times {\text{10}}^{-\text{34}}\text{ J}\cdot \text{s}$, and KE is the kinetic energy of an ejected electron.