A single 218U nucleus, initially at rest, emits an alpha particle as it decays i

Question

A single 218U nucleus, initially at rest, emits an alpha particle as it decays into a 214Th nucleus. There are no other decay products, and the kinetic energy of the alpha particle is measured to be 1.382 × 10-12 J. What is the total amount of nuclear potential energy, in joules, that was released in this process? Answer in Joules

What is the difference between the original mass of the 218U nucleus and the total mass of the fragments (214Th and alpha particle) after the decay? Answer in kilograms

Explanation / Answer

This is a question of nuclear binding and mass defect:

When 218U breaks in 214Th(thorium nucleus the number of protons and neutrons are

np = 90

Nn= 124

To calculate mass defect the formula is:

Md= (Mn+Mp)-Mo

Md=mass defect

Mn=neutorn mass

Mp = proton mass

Mo is observed mass

here total number of protons and neutns will be calculated as:

Total number of thorium protons +total number of protons in the alpha particle = 90+2 = 92

Simliarly,no of neutrons = 124+2 = 126

Mass of one proton = 1.00728 atomic mass unit

Mass of one neutron = 1.00867 amu

Substituting the values in the formula-

Md=(92(1.00728 amu)+126(1.00867 amu))-218

=(92.66976 + 127.08368)- 218

=219.3536-218

Md = 1.75336 amu

Md(iin kgs) = 1.75336 * ( 1/6.02214 * 10^26) = 2.911* 10 ^ 27 Kgs

The energy is calculated as:

E = mc2

=2.911*(10^-27)*(2.9979*10^8)2

=2.60246*(10^-10) Joules

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