Chapter 19, Problem 88IP

Interpretation Introduction

Interpretation: Particle emitted in the given nuclear synthesis is to be stated. Binding energy in $\text{J/nucleon}$ for cobalt- $60$ is to be calculated. De Broglie wavelength of the emitted particle is to be stated.

Concept introduction: A process through which an unstable nuclide looses its energy due to excess of protons or neutrons is known as radioactive decay. Decay constant is the quantity that expresses the rate of decrease of number of atoms of a radioactive element per second. Half life of radioactive sample is defined as the time required for the number of nuclides to reach half of the original value.

To determine: Particle emitted in the given nuclear synthesis, binding energy in $\text{J/nucleon}$ for cobalt- $60$ and De Broglie wavelength of the emitted particle.

Answer to Problem 88IP

Answer

The particle emitted is beta particle.

The binding energy per nucleon is $\underset{\_}{1.402\times 10^{-12}J/nucleon}$.

The de Broglie wavelength is $\underset{\_}{2.6938\times 10^{-13}m}$ .

Explanation of Solution

Explanation

The nuclear reaction for the synthesis of cobalt- $60$ is,

${}_{26}^{58}\text{Fe}+2_0^1\text{n}\to {}_{27}^{60}\text{Co}+{}_{-1}^0\text{e}$.

The synthesis involves the emission of beta particle.

The atomic mass of ${}^{60}\text{Co}=59.9338\text{amu}$

The mass of a neutron is $1.0087\text{amu}$.

The mass of ${}_{\text{1}}^{\text{1}}\text{H}$ proton is $1.00782\text{amu}$.

Number of protons in ${}^{60}\text{Co}=27$.

Number of neutrons in ${}^{60}\text{Co}=33$.

The mass defect is calculated by the formula,

$\Delta m=\text{Atomic}\text{mass of }{}^{60}\text{Co}-\left[\begin{array}{l}\text{Number}\text{of}\text{protons}\times \text{mass}\text{of}{}_{\text{1}}^{\text{1}}\text{H}\text{proton}-\\ \text{Number}\text{of}\text{neutrons}\times \text{mass}\text{of}\text{neutron}\end{array}\right]$

Substitute the value of the atomic mass of ${}^{60}\text{Co}$, the number of protons and mass of the ${}_{\text{1}}^{\text{1}}\text{H}$ proton and that of the neutron in the above equation.

$\begin{array}{l}\Delta m=59.9338-\left[\left(27\times 1.00782\right)+\left(33\times 1.00866\right)\right]\\ \Delta m=-0.56312\text{amu}/\text{nucleus}\end{array}$

The conversion of $\text{amu}/\text{nucleus}$ to $\text{Kg/nucleus}$ is done as,

$1\text{amu}=1.66\times {10}^{-27}\text{Kg}$

Therefore, the conversion of $-0.56444\text{amu}/\text{nucleus}$ into $\text{kg/nucleus}$ is,

$\begin{array}{c}-0.56312\text{amu}/\text{nucleus}=\left(-0.56312\times 1.66\times {10}^{-27}\right)\text{kg/nucleus}\\ =-\text{9}{\text{.347×10}}^{\text{-28}}\text{Kg/nucleus}\end{array}$

Therefore, the mass defect $\left(\Delta m\right)$ of ${}^{60}\text{Co}$ is $-\text{9}{\text{.347×10}}^{\text{-28}}\text{Kg/nucleus}$.

The binding energy per nucleon is calculated by Einstein’s mass energy equation, that is,

$\Delta E=\Delta m\cdot C^2$

Where,

• $\Delta E$ is the change in energy.
• $\Delta m$ is the change in mass.
• $C$ is the velocity of light $\left(3\times {10}^8\right)$.

Substitute the values of $\Delta m$ and $C$ in the equation.

$\begin{array}{l}\Delta E=\Delta m\cdot c^2\\ \Delta E=\left(-9.347\times {10}^{-28}\right){\left(3\times {10}^8\right)}^2\\ \Delta E=8.4123{\text{×10}}^{-\text{11}}\text{J/nucleus}\end{array}$

Therefore, the energy released per nucleus is $8.4123{\text{×10}}^{-\text{11}}\text{J/nucleus}$.

The binding energy per nucleon is calculated by the formula,

$\text{Binding energy per nucleon}=\frac{\text{Binding}\text{energy per nucleus}}{\text{Number}\text{of}\text{protons}+\text{Number}\text{of}\text{neutrons}}$

Substitute the value of the binding energy per nucleus, the number of protons and the number of neutrons in the above equation.

$\begin{array}{c}\text{Binding energy per nucleon}=\frac{\text{Binding}\text{energy per nucleus}}{\left(\text{Number}\text{of}\text{protons}+\text{Number}\text{of}\text{neutrons}\right)}\\ =\frac{8.4123{\text{×10}}^{-\text{11}}}{60}\text{J/nucleon}\\ =\underset{\_}{1.402\times 10^{-12}J/nucleon}\end{array}$

The mass of beta particle is $0.00549\text{amu/nucleon}$.

The conversion of $\text{amu}$ to $\text{Kg}$ is done as,

$1\text{amu}=1.66\times {10}^{-27}\text{Kg}$

Therefore, the conversion of $0.00549\text{amu/nucleon}$ into $\text{kg/nucleon}$ is,

$\begin{array}{c}0.00549\text{amu/nucleon}=\left(0.00549\text{amu/nucleon}\times 1.66\times {10}^{-27}\right)\text{kg/nucleon}\\ =9.11\times {10}^{-30}\text{kg/nucleon}\end{array}$

Therefore, the mass of beta particle is $9.11\times {10}^{-30}\text{kg/nucleon}$.

The de Broglie wavelength is calculated by the formula,

$\lambda =\frac{h}{m\cdot \upsilon }$

Where,

• $h$ is the planks constant.
• $\lambda$ is the de Broglie wavelength.
• $m$ is the mass of the particle.
• $\upsilon$ is the velocity of emitted particle.

The value of $h=$ $6.626\times {10}^{-34}\text{J}\cdot \text{s}$.

The value of $\upsilon$ is $0.90\text{c}$.

The conversion of $\text{c}$ to $\text{m/s}$ is done as,

$1\text{c}=\text{3}\times {\text{10}}^8\text{m/s}$

Therefore, the conversion of $0.90\text{c}$ to $\text{m/s}$ is,

$0.90\text{c}=0.90\times 3\times {10}^8\text{m/s}=\text{27}\times {\text{10}}^7\text{m/s}$

Therefore, the velocity of emitted particle is $\text{27}\times {\text{10}}^7\text{ m/s}$.

Substitute the values of, $h$, $m$ and $\upsilon$ in the above equation.

$\lambda =\frac{6.626\times {10}^{-34}}{9.11\times {10}^{-30}\times \text{27}\times {\text{10}}^7}=\underset{\_}{2.6938\times 10^{-13}m}$

Therefore, the de Broglie wavelength is $\underset{\_}{2.6938\times 10^{-13}m}$.

Conclusion

Conclusion

The particle emitted is beta particle.

The binding energy per nucleon is $\underset{\_}{1.402\times 10^{-12}J/nucleon}$.

The de Broglie wavelength is $\underset{\_}{2.6938\times 10^{-13}m}$.