Chapter 6, Problem 103AP

Interpretation Introduction

Interpretation:

The binding energy of each metal, and the metal with highest binding energy is to be predicted; and the metal thatejects electron when photon is fired on it is to be determined.

Concept introduction:

The energy of a photon can be expressed as follows:

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

Here, E is the energy of photon, $\text{h}$ is Planck’s constant $\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)$, $\text{c}$ is the speed of light $\left(3.0\times {10}^8\text{ m/s}\right)$, and $\lambda$ is the wavelength.

The relationship between wavelength and frequency is as follows:

$\text{c}=\lambda \times \nu$

Here, $\text{c}$ is the speed of light $\left(3.0\times {10}^8\text{ m/s}\right)$, $\lambda$ is the wavelength, and $\nu$ is frequency.

The relationship between binding energy, kinetic energy, and energy of ejected photon is as follows:

$\text{E}=\text{h}\nu =W+\text{K}\text{. E}$

Here, E is the energy of ejected photon, $h$ is Planck’s constant, $\nu$ is frequency of radiation falling on the metal, $W$ is the binding energy, and K. E. is the kinetic energy of the ejected photon.

Answer to Problem 103AP

Solution:

(a)

The binding energy of metals A, B, and C is $3.4\times {10}^{-19}\text{ J}$, $5.6\times {10}^{-19}\text{ J}$, and $6.6\times {10}^{-19}\text{ J}$, respectively. Metal C has the highest binding energy.

(b)

An electron can be ejected when a radiation of minimum frequency falls on the metal surface. Therefore, on allowing a radiation of wavelength 333 nmto fall on metal Ametal B, they will eject electrons.

Explanation of Solution

(a)

Given information: The frequency of metal A $\left({\nu }_A\right)$ is $5.2\times {10}^{14}{\text{ s}}^{-1}$.

The frequency of metal B $\left({\nu }_B\right)$ is $8.5\times {10}^{14}{\text{ s}}^{-1}$.

The frequency of metal C $\left({\nu }_C\right)$ is $9.9\times {10}^{14}{\text{ s}}^{-1}$.

Consider the binding energy for metal A, B, and C tobe $W_A$, $W_B$, and $W_C$, respectively.

In the photoelectric effect, the minimum frequency (or maximum wavelength) that is required to dislodge an electron from the metal surface is called threshold energy. The extra energy is stored in the electron as kinetic energy (K. E.). If kinetic energy becomes zero and the binding energy is equal to the energy of the ejected photon, that is,

$W=\text{h}\nu$

Now, the binding energy of electron for each metal can be evaluated as shown below:

For metal A:

$\begin{array}{c}{W}_A=\text{h}{\nu }_A\\ =\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(5.2\times {10}^{14}{\text{ s}}^{-1}\right)\\ =3.4\times {10}^{-19}\text{ J}\end{array}$

For metal B:

$\begin{array}{c}{W}_B=\text{h}{\nu }_B\\ =\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(8.5\times {10}^{14}{\text{ s}}^{-1}\right)\\ =5.6\times {10}^{-19}\text{ J}\end{array}$

For metal C:

$\begin{array}{c}{W}_C=\text{h}{\nu }_C\\ =\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(9.9\times {10}^{14}{\text{ s}}^{-1}\right)\\ =6.6\times {10}^{-19}\text{ J}\end{array}$

Therefore, the binding energy of metal C is highest.

(b)

Given information:

The wavelength of photon is $333\text{ nm}$.

The frequency of metal A $\left({\nu }_A\right)$ is $5.2\times {10}^{14}{\text{ s}}^{-1}$.

The frequency of metal B $\left({\nu }_B\right)$ is $8.5\times {10}^{14}{\text{ s}}^{-1}$.

The frequency of metal C $\left({\nu }_C\right)$ is $9.9\times {10}^{14}{\text{ s}}^{-1}$.

The frequency associated with $333\text{ nm}$ wavelength can be evaluated as follows:

$\begin{array}{c}\text{c}=\lambda \times \nu \\ 3.0\times {10}^8\text{ m/s}=333\text{ nm}\times \nu \\ \nu =\frac{3.0\times {10}^8\text{ m/s}}{333\text{ nm}}\left(\frac{1\text{ nm}}{1\times {10}^{-9}\text{ m}}\right)\\ =9.0\times {10}^{14}{\text{ s}}^{-1}\end{array}$

So, the frequency is $9.0\times {10}^{14}{\text{ s}}^{-1}$.

Since the electron can be ejected when a radiation of minimum frequency falls on the metal surface;therefore, on allowing a radiation of wavelength $333\text{ nm}$ to fall on metal A and metal B, they will eject electrons.