Chapter 30, Problem 55A

To determine

To Calculate: The mass of helium isotope in atomic mass units.

Answer to Problem 55A

  $\text{4}\text{.27}\text{u}$

Explanation of Solution

Given:

Binding energy of $\text{H}\text{e}$ , $B.E=228.3\text{MeV}$

Formula Used:

B.E = (mass of nucleus − mass of particles) $\times c^2$

Calculations:

Energy equivalent of mass of proton is $938.28\text{MeV}$ .

Energy equivalent of mass of neutron is $939.57\text{MeV}$ .

  $\begin{array}{l}B.E.=m\left(\text{H}\text{e}\right)c^2-2m_p{c}^2-2m_n{c}^2\\ 228.3\text{MeV=}m\left(\text{H}\text{e}\right)c^2-2\left(938.28\text{MeV}\right)-2\left(939.57\text{MeV}\right)\\ m\left(\text{H}\text{e}\right)c^2=3984\text{MeV}\\ \text{1 u = }931.5{\text{MeV/c}}^{\text{2}}\\ m\left(\text{H}\text{e}\right)=\frac{3984{\text{MeV/c}}^2}{931.5{\text{MeV/c}}^{\text{2}}}\text{u}\\ m\left(\text{H}\text{e}\right)\text{= 4}\text{.27}\text{u}\end{array}$

Conclusion:

Thus, mass of Helium isotope is $\text{4}\text{.27}\text{u}$ .