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Mass Energy Equivalence

Find the ener...

Question

Find the energy needed to remove a neutron from the nucleus of the calcium isotope $20^{46} C a$

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Solution

$_{20}^{41} C a$ nucleus is formed removing a neutron from $_{20}^{42} C a$ .
The mass of $_{20}^{41} C a$ plus the mass of a free neutron equals $40.962278 u$
Mass of neutron $= 40.962278 u u + 1.008665 u = 41.970943 u$
Difference between $_{20}^{42} C a$ plus the mass of a free neutron and the mass of $_{20}^{42} C a$ is 0.012321 u.
So, the binding energy of the missing neutron is
$(0.012321 u) (931.49 M e V / u) = 11.5 M e V$

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_{13}^{27} A l

from the following data:

m (_{13}^{26} A l) = 25.986895 u

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, and mass of neutron

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