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Chapter 37: 42P (page 1148)

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

Expert verified

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

Step by step solution

01

Identification of given data

The mass of Carbon nuclei is mc=11.99671u

The mass of Helium nuclei is mHe=4.00151u

The number of nuclei of Helium is 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.

02

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:

E=nmH-mCc2

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.

E=34.00151u-11.99671u931.5MeV1u=7.28MeV

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

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