Question

What is the minimum energy that is required to break a nucleus of ¹²C (of mass 11.996 71 u) into three nuclei of ⁴He (of mass 4.001 51 u each)?

Short answer

The minimum energy to break the nucleus of Carbon nuclei into Helium nuclei is 7.28 MeV.

Step-by-step solution

Identification of given data

The mass of Carbon nuclei is ${\mathrm{m}}_{\mathrm{c}}=11.99671\mathrm{u}$

The mass of Helium nuclei is ${\mathrm{m}}_{\mathrm{He}}=4.00151\mathrm{u}$

The number of nuclei of Helium is $\mathrm{n}=3$

The mass defect is the amount of mass lost in the breaking of nucleus of large atom into small atoms. The lost mass converted into energy.

Determination of minimum energy required to break the Carbon nucleus

The minimum energy to break the nucleus of Carbon nuclei into Helium nuclei is given as:

$\mathrm{E}=\left({\mathrm{nm}}_{\mathrm{H}}-{\mathrm{m}}_{\mathrm{C}}\right){\mathrm{c}}^2$

Here, c is the speed of light and its value in terms of mass and energy is 931.5 MeV/u

Substitute all the values in the above equation.

$\begin{array}{rcl}\mathrm{E}& =& \left[\left(3\right)\left(4.00151\mathrm{u}\right)-11.99671\mathrm{u}\right]\left(\frac{931.5\mathrm{MeV}}{1\mathrm{u}}\right)\\ & =& 7.28\mathrm{MeV}\end{array}$

Therefore, the minimum energy to break the nucleus of Carbon nuclei into Helium nuclei is 7.28 MeV.