Chapter 30.1, Problem 9SSC

(a)

To determine

To calculate: The Mass defect $\left(\Delta m\right)$ .

Answer to Problem 9SSC

The mass of radioactive carbon isotope $C$ is $\text{0}\text{.113196}\text{u}$ .

Explanation of Solution

Given:

Mass of an atom, $m_{atom}=14.003074\text{u}$

Mass of hydrogen $=1.007825\text{u}$

Mass of neutron $=1.008665\text{u}$

  $\text{1}\text{u}\text{=}\text{931}\text{.49}\text{MeV}$

Atomic Mass Number $\left(A\right)=14$

Atomic Number $\left(Z\right)=6$

Formula used:

Mass defect can be calculated by,

  $\Delta m=\left[Zm\left(H\right)+\left(A-Z\right)m_n\right]-m_{atom}$

Where,

  $\Delta m:$ Mass defect

  $m\left(H\right)\text{:}$ Mass of hydrogen

  $m_n\text{:}$ Mass of neutron

  $m_{atom}\text{:}$ Mass of overall atom

  $A:$ Atomic mass number

  $Z:$ Atomic number

Calculation:

To obtain the number of protons,

As the number of protons in the nucleus of the atom is equal to the atomic number $Z$ ,

Number of protons $\text{=6}$

Number of neutrons $=A-Z$

Substitute the values in above formula,

  $\begin{array}{l}=14-6\\ =8\end{array}$

Number of neutrons $=8$

Mass defect can be calculated by formula,

  $\Delta m=\left[Zm\left(H\right)+\left(A-Z\right)m_n\right]-m_{atom}$

Substituting the values in above formula,

  $\begin{array}{l}\Delta m\text{=}\left[\left(\text{6×1}\text{.007825}\text{u}\right)\text{+}\left(\text{8×1}\text{.008665}\text{u}\right)\right]\text{-14}\text{.00307}\text{u}\\ \text{=}\left[\text{6}\text{.04695}\text{u+8}\text{.06932}\text{u}\right]\text{-14}\text{.00307}\text{u}\\ \text{=0}\text{.113196}\text{u}\end{array}$

Conclusion:

The mass of radioactive carbon isotope $C$ is $\text{0}\text{.113196}\text{u}$ .

(b)

To determine

To calculate:The Binding energy.

Answer to Problem 9SSC

The binding energy of the radioactive carbon isotope $C$ is $\text{9}{\text{.4058×10}}^{\text{18}}\text{MeV}$ .

Explanation of Solution

Given:

Mass defect, $\Delta m=0.113196\text{u}$

Formula used:

Formula to calculate binding energy, $E=\Delta m\times c^2$

Where,

  $\Delta m:$ Mass defect

  $c:$ Speed of light

Calculation:

It is given that $1\text{u=}931.49\text{MeV}$

Binding energy in $\text{MeV}$ can be given as,

  $E=\Delta m\times c^2$

Substituting the values in above formula

  $\begin{array}{l}\text{=}\left(\text{0}\text{.112196}\text{u}\right)\text{×}{\left({\text{3×10}}^{\text{8}}\right)}^{\text{2}}\\ \text{=}\left(\text{0}\text{.112196}\text{u}\right)\text{×}\left({\text{9×10}}^{\text{16}}\right)\\ \text{=1}{\text{.009764×10}}^{\text{16}}\text{u}\end{array}$

Converting binding energy into $\text{MeV}$ ,

  $\begin{array}{l}\text{=}\left(\text{1}{\text{.009764×10}}^{\text{16}}\text{u}\right)\text{×}\left(\text{931}\text{.49}\text{MeV}\right)\\ \text{=9}{\text{.4058×10}}^{\text{18}}\text{MeV}\end{array}$

Conclusion:

The binding energy of the radioactive carbon isotope $C$ is $\text{9}{\text{.4058×10}}^{\text{18}}\text{MeV}$ .