Chapter 30, Problem 70GP

(a)

To determine

The mass excess for $\text{H}\text{e}$ in u and ${\text{MeV}}^{\text{2}}{\text{/c}}^{\text{2}}$ .

Explanation of Solution

Given:

Mass number of $\text{H}\text{e}$ is 4.

Atomic mass of $\text{H}\text{e}$ is $4.002603\text{u}$ .

Formula Used:

Mass excess:

  $\Delta =M-A$

Calculations:

  $\begin{array}{c}\Delta =M-A\\ =4.002603\text{u}-4\\ =0.002603\text{u}\\ \text{=}0.002603\text{u}\times \left(\text{931}{\text{.5MeV/uc}}^{\text{2}}\right)\\ =2.42{\text{MeV/c}}^{\text{2}}\end{array}$

Conclusion:

Thus, the mass excess for $\text{H}\text{e}$ is $0.002603\text{u}$ or $2.42{\text{MeV/c}}^{\text{2}}$ .

(b)

To determine

The mass excess for $\text{C}$ in u and ${\text{MeV}}^{\text{2}}{\text{/c}}^{\text{2}}$ .

Explanation of Solution

Given:

Mass number of $\text{C}$ is 12.

Atomic mass of $\text{C}$ is $12.0000\text{u}$ .

Formula Used:

Mass excess:

  $\Delta =M-A$

Calculations:

  $\begin{array}{c}\Delta =M-A\\ =12.0000\text{u}-12\\ =0.00\text{u}\\ =0.00{\text{MeV/c}}^{\text{2}}\end{array}$

Conclusion:

Thus, the mass excess for $\text{C}$ is $0.00\text{u}$ or $0.00{\text{MeV/c}}^{\text{2}}$ .

(c)

To determine

The mass excess for $\text{A}\text{g}$ in u and ${\text{MeV}}^{\text{2}}{\text{/c}}^{\text{2}}$ .

Explanation of Solution

Given:

Mass number of $\text{A}\text{g}$ is 107.

Atomic mass of $\text{A}\text{g}$ is $106.905093\text{u}$ .

Formula Used:

Mass excess:

  $\Delta =M-A$

Calculations:

  $\begin{array}{c}\Delta =M-A\\ =106.905093\text{u}-107\\ =-0.094907\text{u}\\ =-88.4{\text{MeV/c}}^{\text{2}}\end{array}$

Conclusion:

Thus, the mass excess for $\text{A}\text{g}$ is $-0.094907\text{u}$ or $-88.4{\text{MeV/c}}^{\text{2}}$ .

(d)

To determine

The mass excess for $\text{U}$ in u and ${\text{MeV}}^{\text{2}}{\text{/c}}^{\text{2}}$ .

Explanation of Solution

Given:

Mass number of $\text{U}$ is 235.

Atomic mass of $\text{U}$ is $235.043923\text{u}$ .

Formula Used:

Mass excess:

  $\Delta =M-A$

Calculations:

  $\begin{array}{c}\Delta =M-A\\ =235.043923\text{u}-235\\ =0.043923\text{u}\\ =40.9{\text{MeV/c}}^{\text{2}}\end{array}$

Conclusion:

Thus, the mass excess for $\text{U}$ is $0.043923\text{u}$ or $40.9{\text{MeV/c}}^{\text{2}}$ .

(e)

To determine

Whether sign of $\Delta$ depends on A or Z.

Explanation of Solution

Introduction:

Mass excess:

  $\Delta =M-A$

Here, M is atomic mass and A is the mass number.

In general, the sign of $\Delta$ depends on A which is sum of Z and number of neutrons.