Examples

Section 7.3 Examples

Example 7.3.1.

Determine the number of protons, electrons, and neutrons in the \({}_{92}^{236}U\) atom.

Solution.

From isotopic notation (7.2.1). we have atomic number, Z = 92. So the nuber of protons (= number of electrons) = 92. The mass number, A =236. Hence the nuber of neutrons, \(N=A-Z=236-92= 144.\)

Example 7.3.2.

Write the equation for one alpha decay in \({}_{92}^{238}U\text{.}\)

Solution.

\begin{equation*} {}_{92}^{238}U \rightarrow {}_{2}^{4}He + {}_{90}^{234}X \end{equation*}

Here, X stands for the element of atomic number 90, which is \(Th\text{.}\)

Example 7.3.3.

If 1.0g of caesium (Cs-137) disintegrates over a period of 90 years. How many grams of Cs-137 would remain, if Cs-137 has half-life of 30 years.

Solution.

For 90 years, we have to wait for 3 half-life of Cs-137. Every half-life it reduces 1/2 of of its amount. Now from (7.2.2), we have \(\left(\frac{1}{2}\right)^3 \times 1 \,g =0.125 \,g. \)

Example 7.3.4.

How long does it take a 100.00g sample of As-81, with a half-life of 33 seconds, to decay to 6.25 g?

Answer.

4 half-life = \(4\times 33\) second = 132 seconds.

Solution.

In \(1^{st}\) HL remaining amount is 50 g, In \(2^{nd}\) HL remaining amount is 25 g, In \(3^{rd}\) HL remaining amount is 12.5 g, and in \(4^{th}\) HL the remaining amount is 6.25 g.

Example 7.3.5.

Complete the equation if proton \({}_{1}^{1}H\) strikes on the \({}_{80}^{200}Hg\text{.}\)

\begin{equation*} {}_{1}^{1}H + {}_{80}^{200}Hg \rightarrow \cdots + {}_{2}^{4}He \end{equation*}

Solution.

\begin{equation*} {}_{1}^{1}H + {}_{80}^{200}Hg \rightarrow {}_{79}^{197}X + {}_{2}^{4}He \end{equation*}

\(\therefore \quad X=Au\)

Example 7.3.6.

Complete the following fission reaction.

\begin{equation*} {}_{92}^{236}U \rightarrow {}_{38}^{90}Sr + \cdots + 2\cdot {}_{0}^{1}n \end{equation*}

Example 7.3.7.

Calculate the binding energy in the following fission reaction.

\begin{equation*} {}_{92}^{236}U \rightarrow {}_{36}^{88}Kr + {}_{56}^{144}Ba+ 4\cdot {}_{0}^{1}n \end{equation*}

The mass of \({}_{92}^{236}U\) is \(236.04556 \,u\text{,}\) \({}_{36}^{88}Kr\) is \(87.91445 \,u\text{,}\) \({}_{56}^{144}Ba\) is \(143.92284 \,u\text{,}\) and that of \({}_{0}^{1}n\) is \(1.00867 \,u\text{.}\)

Solution.

The total mass on the left side of the arrow is \(236.04556 u\) and the maas on the right side of arrow is \(235.87197 \,u\text{.}\) Hence the mass defect is \(0.17359 \,u\text{.}\) Now from Subsection 7.2.2, we have \(1\,u = 931 \,MeV.\) Therefore the binding energy is \(0.17359 \,u\times 931 \,MeV/u = 162 MeV.\)

Example 7.3.8.

If \((\frac{7}{8})^{th}\) of the sample is decayed in 15 days. What is its half-life?

Solution.

From (7.2.2)

\begin{align*} N \amp = N_o\left(\frac{1}{2}\right)^n \end{align*}

Here, \(N=N_o-\frac{7}{8}(N_o) = \frac{1}{8}N_o\text{.}\) Suppose number of half-life, \(n=t/T\text{,}\) where \(t\) is given time period and \(T\) is half-life period, then

\begin{align*} \frac{1}{8}N_o \amp = N_o\left(\frac{1}{2}\right)^{\frac{t}{T}}\\ or, \quad \frac{1}{8} \amp = \left(\frac{1}{2}\right)^{\frac{15}{T}}\\ or, \quad \left(\frac{1}{2}\right)^3 \amp = \left(\frac{1}{2}\right)^{\frac{15}{T}}\\ or, \quad 3 \amp = \frac{15}{T}\\ \therefore T \amp = \frac{15}{3} =5 days \end{align*}

Example 7.3.9.

If \((\frac{1}{16})^{th}\) of the sample is decayed in 2 hours. What is its half-life?

Solution.

If \((\frac{1}{2})\) of the sample decayes in every 30 minutes then the sample must be decayed by (1/16)^th in 2 hours. Hence 30 minute is a half-life. \(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}= \frac{1}{16}\)

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