Chapter 1, Problem 1B.15E

(a)

Interpretation Introduction

Interpretation:

The wavelength of the ejected electron having wavelength of $3.6\times {10}^{3}km.s^{-1}$ has to be calculated.

Concept Introduction:

de Broglie relation:

Mathematically, de Broglie relation can be represented as given below.

  $\lambda =\frac{h}{mv}$

Where, $h$ is the Planck’s constant, $\lambda$ is the wavelength of radiation, m is the mass of particle and v is the velocity of the particle.

Answer to Problem 1B.15E

The wavelength of the ejected electron having wavelength of $3.6\times {10}^{3}km.s^{-1}$ is $2.02\times 10^{-10}m.$

Explanation of Solution

Given that, the velocity of electron is $3.6\times {10}^{3}km.s^{-1}.$  The mass of electron is $9.109\times {10}^{-31}kg.$

The de Broglie relation is given below.

  $\lambda =\frac{h}{mv}$

By plugging all data in the above equation, the value of wavelength of the electron can be calculated.

  $\begin{array}{l}\lambda =\frac{h}{mv}=\frac{6.626\times {10}^{-34}J.s}{\left(9.109\times {10}^{-31}kg\right)\times \left(3.6\times {10}^{6}m/s\right)}\\ \\ \lambda =2.02\times {10}^{-10}\times \frac{kg.m^2.s^{-2}.s}{kg.m.s^{-1}}\\ \\ \lambda =2.02\times 10^{-10}m\end{array}$

Therefore, the wavelength of the ejected electron having wavelength of $3.6\times {10}^{3}km.s^{-1}$ is $2.02\times 10^{-10}m.$

(b)

Interpretation Introduction

Interpretation:

The energy required to remove the electron from the metal has to be calculated.

Concept Introduction:

Energy of a photon can be expressed mathematically as given below.

  $E=h\nu$

Where, h is the Planck’s constant, $\nu$ is the frequency of radiation.

Answer to Problem 1B.15E

The energy required to remove the electron from the metal is $1.66\times 10^{-17}J.$

Explanation of Solution

Given that, the frequency of the radiation is $2.50\times {10}^{16}Hz.$  The corresponding energy of a photon can be calculated as given below.

  $E=h\nu =\left(6.626\times {10}^{-34}J.s\right)\times \left(2.50\times {10}^{16}{s}^{-1}\right)=1.66\times 10^{-17}J.$

Therefore, the energy required to remove the electron from the metal is $1.66\times 10^{-17}J.$

(c)

Interpretation Introduction

Interpretation:

The wavelength of the radiation that caused photoejection of electron with a velocity of $3.6\times {10}^{3}km.s^{-1}$ has to be calculated.

Concept Introduction:

Photoelectric effect:

If the energy of the photon is greater than work function, then an electron can be ejected with a kinetic energy, $E_k=\frac{1}{2}{m}_e{v}^2.$  An equation which represents the kinetic energy is given below.

  $E_k=h\nu -\Phi$

Where, $E_k$ is the kinetic energy, $h$ is the Planck’s constant, $\nu$ is the frequency of radiation and $\Phi$ is the work function.

Answer to Problem 1B.15E

The wavelength of the radiation that caused photoejection of electron is $8.8nm.$

Explanation of Solution

Given that, the speed of an electron that is emitted from the surface of a sample of chromium metal by a photon is $3.6\times {10}^{3}km.s^{-1}.$

The expression of kinetic energy is given below.

  $E_k=\frac{1}{2}{m}_e{v}^2\mathrm{.......}\left(1\right)$

Where, $m_e$ is the mass of electron and $v$ is the speed of electron.

By plugging all data in the above equation, the value of kinetic energy can be calculated.

  $\begin{array}{c}{E}_k=\frac{1}{2}{m}_e{v}^2\\ \\ E_k=\frac{1}{2}\times \left(9.109\times {10}^{-31}kg\right)\times {\left(3.6\times {10}^{6}m.s^{-1}\right)}^2\\ \\ {\text{E}}_{\text{k}}\text{=}\text{5}{\text{.9×10}}^{\text{-18}}\text{kg}{\text{.m}}^{\text{2}}{\text{.s}}^{\text{-2}}=5.9\times 10^{-18}J.\end{array}$

The energy of incoming photon can be calculated as given below.

  $\begin{array}{l}E\left(\text{incoming}\text{photon}\right)=E_k+h\nu \mathrm{..........}\left(2\right)\\ \\ \left[\because \text{}\Phi =h\nu \right]\end{array}$

By plugging all data in the above equation, the energy of incoming photon can be calculated.

  $\begin{array}{l}E\left(\text{incoming}\text{photon}\right)=E_k+h\nu \\ \\ E\left(\text{incoming}\text{photon}\right)=E_k+h\nu =\left(\text{5}{\text{.9×10}}^{\text{-18}}\text{J}\right)\text{+}\left(1.66\times {10}^{-17}J\right)\\ \\ E\left(\text{incoming}\text{photon}\right)=2.25{\text{×10}}^{\text{-17}}\text{J}\\ \\ \frac{hc}{\lambda }=2.25{\text{×10}}^{\text{-17}}\text{J}\\ \\ \lambda =\frac{hc}{2.25{\text{×10}}^{\text{-17}}\text{J}}=\frac{\left(6.626\times {10}^{-34}J.s\right)\times \left(3\times {10}^{8}m/s\right)}{2.25{\text{×10}}^{\text{-17}}\text{J}}=8.8\times {10}^{-9}m=8.8nm.\end{array}$

Therefore, the wavelength of the radiation that caused photoejection of electron is $8.8nm.$

(d)

Interpretation Introduction

Interpretation:

The kind of electromagnetic radiation used has to be determined.

Concept Introduction:

The wavelengths of electromagnetic radiation and their corresponding frequencies are given below in the table.

Radiation type	Frequency / $\left({10}^{14}Hz\right)$	Wavelength / $\left(nm\right)$	Energy of photon / $\left({10}^{-19}J\right)$
x-rays and $\gamma$- rays	$\ge {10}^3$	$\le 3$	$\ge {10}^3$
Ultraviolet	$8.6$	$350$	$5.7$
Visible light
Violet	$7.1$	$420$	$4.7$
Blue	$6.4$	$470$	$4.2$
Green	$5.7$	$530$	$3.8$
Yellow	$5.2$	$580$	$3.4$
Orange	$4.8$	$620$	$3.2$
Red	$4.3$	$700$	$2.8$
Infrared	$3.0$	$1000$	$2.0$
Microwaves and radio waves	$\le {10}^{-3}$	$\ge 3\times {10}^6$	$\le {10}^3$

Explanation of Solution

The wavelength of the radiation that caused photoejection of electron is $8.8nm.$

Therefore, the kind of electromagnetic radiation used is X- rays.