Answer:
c. 2 MeV.
Explanation:
The computation of the binding energy is shown below
![= [Zm_p + (A - Z)m_n - N]c^2\\\\=[(1) (1.007825u) + (2 - 1 ) ( 1.008665 u) - 2.014102 u]c^2\\\\= (0.002388u)c^2\\\\= (.002388) (931.5 MeV)\\\\=2.22 MeV](https://tex.z-dn.net/?f=%3D%20%5BZm_p%20%2B%20%28A%20-%20Z%29m_n%20-%20N%5Dc%5E2%5C%5C%5C%5C%3D%5B%281%29%20%281.007825u%29%20%2B%20%282%20-%201%20%29%20%28%201.008665%20u%29%20-%202.014102%20u%5Dc%5E2%5C%5C%5C%5C%3D%20%280.002388u%29c%5E2%5C%5C%5C%5C%3D%20%28.002388%29%20%28931.5%20MeV%29%5C%5C%5C%5C%3D2.22%20MeV)
= 2 MeV
As 1 MeV = (1 u) c^2
hence, the binding energy is 2 MeV
Therefore the correct option is c.
We simply applied the above formula so that the correct binding energy could come
And, the same is to be considered
The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.
<h3>How does test charge affect electric field?</h3>
As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.
Adjusting the amount of charge on the test charge will not change the electric field force.
<h3>What is a test charge used for?</h3>
The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.
To learn more about test charge, refer
brainly.com/question/16737526
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Answer:
In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
