Chapter 1, Problem 1.31E

(a)

Interpretation Introduction

Interpretation:

Laser that can be used has to be determined among a high-intensity red ruby laser and a low-intensity violet GaN laser.

Concept Introduction:

Wavelength and frequency are inversely proportional to each other and the relationship between wavelength and frequency can be given as,

  $\text{ν}\text{=}\frac{\text{c}}{\text{λ}}$

Here, $\text{ν}$ is the frequency, $\text{λ}$ is the wavelength and c is the speed of light

Planck’s equation,

    $\text{E = hν}$

Here, $\text{ν}$ is the frequency

Explanation of Solution

• Calculate the energy of light from a high-intensity red ruby laser:

Wavelength of light from a high-intensity red ruby laser is $\text{694 nm = 694}\times {\text{10}}^{-9}m$.

The frequency of light from a high-intensity red ruby laser is calculated by using the equation,

    $\begin{array}{c}\text{ν}\text{=}\frac{\text{c}}{\text{λ}}\\ \\ =\frac{\text{2}\text{.998}\text{×}\text{10}{}^{\text{8}}\text{m/s}}{\text{694}\times {\text{10}}^{-9}m}\\ \\ =4.32\text{×}\text{10}{}^{14}{\text{s}}^{-1}\end{array}$

The frequency of light is $4.32\text{×}\text{10}{}^{14}{\text{s}}^{-1}$

The energy per photon of light is calculated,

  $\text{E}\text{=}\text{hν}$

Substituting the values to the above equation,

  $\begin{array}{c}\text{E}\text{=}\text{hν}\\ \\ =\text{6}\text{.626}\text{×}{\text{10}}^{-\text{34}}\text{J}\text{.s}\times 4.32\text{×}\text{10}{}^{14}{\text{s}}^{-1}\\ \\ =2.86\times {10}^{-19}\text{J}\end{array}$

The energy per photon is $2.86\times {10}^{-19}\text{J}$.

• Calculate the energy of light from a low-intensity violet GaN laser:

Wavelength of light from a low-intensity violet GaN laser is $\text{405 nm = 405}\times {\text{10}}^{-9}m$.

The frequency of light from a low-intensity violet GaN laser is calculated by using the equation,

    $\begin{array}{c}\text{ν}\text{=}\frac{\text{c}}{\text{λ}}\\ \\ =\frac{\text{2}\text{.998}\text{×}\text{10}{}^{\text{8}}\text{m/s}}{\text{405}\times {\text{10}}^{-9}m}\\ \\ =7.40\text{×}\text{10}{}^{14}{\text{s}}^{-1}\end{array}$

The frequency of light is $7.40\text{×}\text{10}{}^{14}{\text{s}}^{-1}$

The energy per photon of light is calculated,

  $\text{E}\text{=}\text{hν}$

Substituting the values to the above equation,

  $\begin{array}{c}\text{E}\text{=}\text{hν}\\ \\ =\text{6}\text{.626}\text{×}{\text{10}}^{-\text{34}}\text{J}\text{.s}\times 7.40\text{×}\text{10}{}^{14}{\text{s}}^{-1}\\ \\ =4.90\times {10}^{-19}\text{J}\end{array}$

The energy per photon is $4.90\times {10}^{-19}\text{J}$.

Comparing the values it is clear that the energy of light from a low-intensity violet GaN laser is higher than that of a high-intensity red ruby laser. Hence, a low-intensity violet GaN laser is used since it will give enough energy for ejecting an electron.

(b)

Interpretation Introduction

Interpretation:

Kinetic energy of the electrons emitted has to be calculated.

Concept Introduction:

Photoelectric effect: It is a process where electrons get ejected when a ray of light having adequately high frequency hits on the surface of metal.

Kinetic energy of one ejected electron can be calculated using the given formula,

  $\frac{1}{2}{m}_e{\nu }^2\text{ = h}\nu -\Phi$

Here, $\nu$ is the frequency of incident radiation.

$\frac{1}{2}{m}_e{\nu }^2$ is the kinetic energy of one ejected electron

$\text{h}\nu$ is the energy of the photons in the incident radiation.

$\Phi$ is the energy needed for ejecting an electron (work function).

Answer to Problem 1.31E

Kinetic energy of the electrons emitted is $\text{2}\text{.10}\times {\text{10}}^{-20}\text{ J}$.

Explanation of Solution

Given information is shown below,

  $\begin{array}{c}\Phi \text{ = 2}\text{.93 eV}\\ \\ \text{= 2}\text{.93 eV}\times \frac{\text{1}\text{.602}\times {\text{10}}^{-\text{19}}\text{ J}}{1\text{ eV}}\text{ = 4}\text{.69}\times {\text{10}}^{-19}\text{ J}\end{array}$

Here, a low-intensity violet GaN laser is used

The frequency of light from a low-intensity violet GaN laser is calculated by using the equation,

    $\begin{array}{c}\text{ν}\text{=}\frac{\text{c}}{\text{λ}}\\ \\ =\frac{\text{2}\text{.998}\text{×}\text{10}{}^{\text{8}}\text{m/s}}{\text{405}\times {\text{10}}^{-9}m}\\ \\ =7.40\text{×}\text{10}{}^{14}{\text{s}}^{-1}\end{array}$

The energy per photon of light is calculated,

  $\begin{array}{c}\text{E}\text{=}\text{hν}\\ \\ =\text{6}\text{.626}\text{×}{\text{10}}^{-\text{34}}\text{J}\text{.s}\times 7.40\text{×}\text{10}{}^{14}{\text{s}}^{-1}\\ \\ =4.90\times {10}^{-19}\text{J}\end{array}$

The kinetic energy of the electrons emitted can be calculated as follows,

  $\begin{array}{c}\frac{1}{2}{m}_e{\nu }^2\text{ = h}\nu -\Phi \\ \\ =\text{}4.90\times {10}^{-19}\text{J}-\text{4}\text{.69}\times {\text{10}}^{-19}\text{ J}\\ \\ =\text{ 2}\text{.10}\times {\text{10}}^{-20}\text{ J}\end{array}$

Therefore, kinetic energy of the electrons emitted is $\text{2}\text{.10}\times {\text{10}}^{-20}\text{ J}$.