Chapter 30.1, Problem 7PP

(a)

To determine

To calculate:Mass of the atom $\left(m_{atom}\right)$ .

Answer to Problem 7PP

The mass of nitrogen isotope is $15.0101\text{u}$ .

Explanation of Solution

Given:

Mass defect $\Delta m\text{ =-0}\text{.113986}$

Mass of hydrogen $\text{=1}\text{.007825}\text{u}$

Mass of neutron $\text{=1}\text{.008665}\text{u}$

  $1\text{u = 931}\text{.49}\text{Me}\text{}\text{V}$

Atomic mass number, $A=15$

Atomic number, $Z=7$

Formula used:

Formula to calculate mass defect, $\Delta m=\left[Zm\left(H\right)+\left(A-Z\right)m_n\right]-m_{atom}$

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 $\text{=7}$

Number of neutrons $=A-Z$ ,

Substitute the values in above formula,

  $\begin{array}{l}=15-7\\ =8\end{array}$

Number of neutrons $\text{=8}$

Mass defect can be calculated by formula,

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

Rearranging the above formula to get $m_{atom}$ ,

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

Substituting the values in above formula,

  $\begin{array}{l}\text{=}\left[\left(\text{7×1}\text{.007825}\text{u}\right)\text{+}\left(\text{8×1}\text{.008665}\text{u}\right)\right]\text{- 0}\text{.113986}\\ \text{=}\left[\text{7}\text{.054775}\text{u + 8}\text{.06932}\text{u}\right]\text{ - 0}\text{.113986}\\ \text{=15}\text{.124095}\text{u - 0}\text{.113986}\\ \text{=15}\text{.0101}\text{u}\end{array}$

Conclusion:

The mass of nitrogen isotope is $\text{15}\text{.0101}\text{u}$ .

(b)

To determine

To calculate:The Binding energy.

Answer to Problem 7PP

The binding energy of the nitrogen isotope is $\text{-9}{\text{.5654×10}}^{\text{18}}\text{MeV}$ .

Explanation of Solution

Given:

Mass defect, $\Delta m=\text{-0}\text{.113986}$

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

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

  $E=\Delta m\times c^2$

Substituting the values in above formula

  $\begin{array}{l}\text{=}\left(\text{-0}\text{.113986}\right)\text{×}{\left({\text{3×10}}^{\text{8}}\right)}^{\text{2}}\\ \text{=}\left(\text{-0}\text{.113986}\right)\text{×}\left({\text{9×10}}^{\text{16}}\right)\\ \text{=-1}{\text{.0258×10}}^{\text{16}}\text{u}\end{array}$

Converting the binding energy in $\text{MeV}$ ,

  $\begin{array}{l}\text{=}\left(\text{-1}{\text{.0258×10}}^{\text{16}}\right)\text{×}\left(\text{931}\text{.49}\text{MeV}\right)\\ \text{=-9}{\text{.5654×10}}^{\text{18}}\text{MeV}\end{array}$

Conclusion:

The binding energy of the nitrogen isotope is $\text{-9}{\text{.5654×10}}^{\text{18}}\text{MeV}$ .