Chapter 10, Problem 10A.2E

(a)

Interpretation Introduction

Interpretation:

The frequency and the wavelength of Iron-53 have to be given.

Concept Introduction:

Planck equation:

Planck equation states that the frequency and energy of an electromagnetic radiation are proportional.

  $\text{E}\text{=}\text{hγ}$

Where, E is the energy of the radiation.

h is the planck constant ($\text{6}{\text{.626×10}}^{\text{-34}}\text{Js}$).

$\text{γ}$ is the frequency of the radiation.

Wavelength:

In linear media, any wave pattern can be described in terms of the independent propagation of sinusoidal components.  The wavelength $\left(\text{λ}\right)$ of a sinusoidal waveform travelling at constant speed $\text{c}$ is given by,

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

Where, c is the speed of the light ($\text{2}{\text{.998×10}}^{\text{8}}{\text{ms}}^{\text{-1}}$ ).

$\text{γ}$ is the frequency of the radiation.

Answer to Problem 10A.2E

The frequency and the wavelength of Iron-53 are $\text{0}{\text{.7349×10}}^{\text{21}}\text{s}\text{\&}\text{4}{\text{.07×10}}^{\text{-13}}\text{m}$ respectively.

Explanation of Solution

Given,

The Energy of the radiation (E) =$\text{3}\text{.04Mev}$

$\text{1Mev=1}{\text{.602×10}}^{\text{-13}}\text{J}$

The frequency can be calculated as,

  $\text{γ=}\frac{\text{E}}{\text{h}}$

  $\begin{array}{l}\text{γ=}\frac{\text{3}\text{.04×1}{\text{.602×10}}^{\text{-13}}\text{J}}{\text{6}{\text{.626×10}}^{\text{-34}}\text{Js}}\\ \text{γ=0}{\text{.7349×10}}^{\text{21}}\text{s}\end{array}$

The wavelength can be calculated as,

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

  $\begin{array}{l}\text{λ=}\frac{\text{2}{\text{.998×10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{\text{0}{\text{.7349×10}}^{\text{21}}{\text{s}}^{\text{-1}}}\\ \text{λ=4}{\text{.07×10}}^{\text{-13}}\text{m}\text{.}\end{array}$

The frequency and the wavelength of Iron-53 are $\text{0}{\text{.7349×10}}^{\text{21}}\text{s}\text{\&}\text{4}{\text{.07×10}}^{\text{-13}}\text{m}$ respectively.

(b)

Interpretation Introduction

Interpretation:

The frequency and the wavelength Vanadium-52 has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10A.2E

The frequency and the wavelength of Vanadium are $\text{0}{\text{.3457×10}}^{\text{21}}\text{s}\text{\&}\text{8}{\text{.67×10}}^{\text{-13}}\text{m}$ respectively.

Explanation of Solution

Given,

The Energy of the radiation (E) =$\text{1}\text{.43Mev}$

$\text{1Mev=1}{\text{.602×10}}^{\text{-13}}\text{J}$

The frequency can be calculated as,

  $\text{γ=}\frac{\text{E}}{\text{h}}$

  $\begin{array}{l}\text{γ=}\frac{\text{1}\text{.43×1}{\text{.602×10}}^{\text{-13}}\text{J}}{\text{6}{\text{.626×10}}^{\text{-34}}\text{Js}}\\ \text{γ=0}{\text{.3457×10}}^{\text{21}}\text{s}\end{array}$

The wavelength can be calculated as,

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

  $\begin{array}{l}\text{λ=}\frac{\text{2}{\text{.998×10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{\text{0}{\text{.3457×10}}^{\text{21}}{\text{s}}^{\text{-1}}}\\ \text{λ=8}{\text{.67×10}}^{\text{-13}}\text{m}\text{.}\end{array}$

The frequency and the wavelength of Arsenic-80 are $\text{0}{\text{.3457×10}}^{\text{21}}\text{s}\text{\&}\text{8}{\text{.67×10}}^{\text{-13}}\text{m}$ respectively.

(c)

Interpretation Introduction

Interpretation:

The frequency and the wavelength Scandium-44 has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 10A.2E

The frequency and the wavelength of scandium-44 are $\text{0}{\text{.0580×10}}^{\text{21}}\text{s}\text{\&}\text{51}{\text{.68×10}}^{\text{-13}}\text{m}$ respectively.

Explanation of Solution

Given,

The Energy of the radiation (E) =$\text{0}\text{.27Mev}$

$\text{1Mev=1}{\text{.602×10}}^{\text{-13}}\text{J}$

The frequency can be calculated as,

  $\text{γ=}\frac{\text{E}}{\text{h}}$

  $\begin{array}{l}\text{γ=}\frac{\text{0}\text{.27×1}{\text{.602×10}}^{\text{-13}}\text{J}}{\text{6}{\text{.626×10}}^{\text{-34}}\text{Js}}\\ \text{γ=0}{\text{.0580×10}}^{\text{21}}\text{s}\end{array}$

The wavelength can be calculated as,

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

  $\begin{array}{l}\text{λ=}\frac{\text{2}{\text{.998×10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{\text{0}{\text{.0580×10}}^{\text{21}}{\text{s}}^{\text{-1}}}\\ \text{λ=51}{\text{.68×10}}^{\text{-13}}\text{m}\text{.}\end{array}$

The frequency and the wavelength of Scandium-44 are $\text{0}{\text{.0580×10}}^{\text{21}}\text{s}\text{\&}\text{51}{\text{.68×10}}^{\text{-13}}\text{m}$ respectively.