Chapter 5, Problem 5.14P

(a)

To determine

To plot:

The data given in the table and suggest the type of radiation based on curve.

Answer to Problem 5.14P

Beta and gamma radiation

Explanation of Solution

Calculation:

The graph is plotted for counting per min to the thickness of the absorber. The count for beta particle decreases rapidly and then after certain range it decreases slowly. The counts for beta particle approach zero. But, in the above diagram the count is not approaching zero because of gamma radiations.

Thus, the types of radiation according to the above curve is Beta and gamma radiation.

Conclusion:

The type of radiation suggested by the curve is Beta radiation.

(b)

To determine

The energy of the beta radiations.

Answer to Problem 5.14P

The energy of the beta radiations is, $E=0.985\text{MeV}$

Explanation of Solution

Given:

Range of beta particle, $R=0.4{\text{g/cm}}^2$

Formula used:

The Energy of beta particle is,

  $E=0.85E+0.245$

Where,

  $E$ =Energy of beta particle

Calculation:

From the graph plotted between the thickness and the count per min curve. The range can be estimated as,

  $R=0.4{\text{g/cm}}^2$

The Energy of beta particle is,

  $\begin{array}{l}E=0.85E+0.245\\ E=0.985\text{MeV}\end{array}$

Conclusion:

From the above graph the energy of beta radiation is $E=0.985\text{MeV}$

(c)

To determine

Whether gamma rays are present, the energy of gamma rays if present.

Answer to Problem 5.14P

Yes, the energy is 1.5 MeV.

Explanation of Solution

The attenuation of gamma ray can be described as,

  $\begin{array}{l}I=I_0{e}^{-\mu x}\\ \therefore \mu =\frac{1}{x}\ln \left(\frac{I_0}{I}\right)\end{array}$

As intensity observed initially is 1000. So,

  $I_0=1000$

For $x=0.2\text{mm= 2}\times {\text{10}}^{-4}\text{m}$

  $\begin{array}{l}\mu =\frac{1}{0.02}\ln \left(\frac{1000}{576}\right)\\ \mu =27.58{\text{cm}}^{-1}\end{array}$

For $x=0.4\text{mm= 4}\times {\text{10}}^{-2}\text{cm}$

  $\begin{array}{l}\mu =\frac{1}{4\times {10}^{-2}}\ln \left(\frac{1000}{348}\right)\\ \mu =26.38{\text{cm}}^{-1}\end{array}$

For $x=0.6\text{mm= 6}\times {\text{10}}^{-2}\text{cm}$

  $\begin{array}{l}\mu =\frac{1}{6\times {10}^{-2}}\ln \left(\frac{1000}{230}\right)\\ \mu =24.49{\text{cm}}^{-1}\end{array}$

For $x=0.8\text{mm= 8}\times {\text{10}}^{-2}\text{cm}$

  $\begin{array}{l}\mu =\frac{1}{8\times {10}^{-2}}\ln \left(\frac{1000}{168}\right)\\ \mu =22.29{\text{cm}}^{-1}\end{array}$

For $x=1\text{mm= 0}\text{.1cm}$

  $\begin{array}{l}\mu =\frac{1}{0.1}\ln \left(\frac{1000}{134}\right)\\ \mu =20.09{\text{cm}}^{-1}\end{array}$

For $x=1.2\text{mm=0}\text{.12m}$

  $\begin{array}{l}\mu =\frac{1}{0.12}\ln \left(\frac{1000}{120}\right)\\ \mu =17.66{\text{cm}}^{-1}\end{array}$

For $x=1.4\text{mm= 0}\text{.14cm}$

  $\begin{array}{l}\mu =\frac{1}{0.14}\ln \left(\frac{1000}{107}\right)\\ \mu =15.96{\text{cm}}^{-1}\end{array}$

For $x=1.6\text{mm= }0.16\text{m}$

  $\begin{array}{l}\mu =\frac{1}{0.16}\ln \left(\frac{1000}{96}\right)\\ \mu =14.64{\text{m}}^{-1}\end{array}$

For $x=2\text{mm= }0.2\text{m}$

  $\begin{array}{l}\mu =\frac{1}{0.2}\ln \left(\frac{1000}{95}\right)\\ \mu =11.76{\text{cm}}^{-1}\end{array}$

For $x=4\text{mm= }0.4\text{m}$

  $\begin{array}{l}\mu =\frac{1}{0.4}\ln \left(\frac{1000}{90}\right)\\ \mu =6.01{\text{m}}^{-1}\end{array}$

For $x=8\text{mm= }0.8\text{cm}$

  $\begin{array}{l}\mu =\frac{1}{0.8}\ln \left(\frac{1000}{82}\right)\\ \mu =3.12{\text{cm}}^{-1}\end{array}$

For $x=15\text{mm= }1.5\text{cm}$

  $\begin{array}{l}\mu =\frac{1}{1.5}\ln \left(\frac{1000}{68}\right)\\ \mu =1.79{\text{cm}}^{-1}\end{array}$

For $x=20\text{mm= 2cm}$

  $\begin{array}{l}\mu =\frac{1}{2}\ln \left(\frac{1000}{60}\right)\\ \mu =1.406{\text{cm}}^{-1}\end{array}$

For $x=280\text{mm= 2}\text{.8cm}$

  $\begin{array}{l}\mu =\frac{1}{2.8}\ln \left(\frac{1000}{50}\right)\\ \mu =1.06{\text{cm}}^{-1}\end{array}$

So, mean is,

  $\mu =13.78{\text{cm}}^{-1}$

Thus, the energy corresponding to it is 1.5 MeV.

(d)

To determine

The isotope compatible with absorption data.

Answer to Problem 5.14P

The isotope compatible with above absorption data is ¹⁹⁸Au.

Explanation of Solution

The isotope compatible with above absorption data is ¹⁹⁸Au.

(e)

To determine

The equation fits the absorption data.

Answer to Problem 5.14P

  $I=900e^{-24t}+100e^{-0.25t}$

Explanation of Solution

The equation fits the absorption data is,

  $\begin{array}{l}I=I_0{e}^{-\sigma x}\\ I=900e^{-24t}+100e^{-0.25t}\end{array}$