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What is the nuclear binding energy of an atom that has a mass defect of 1.643 10–28 kg? Use E = mc2. (Remember: The speed of light is approximately 3.00 108 m/s.) 1.83 10–46 J 4.93 10–20 J 1.48 10–11 J 5.48 1045 J

Respuesta :

The energy which is required to split a nucleus into its discrete neutrons and protons is said to be nuclear binding energy.

Using the formula for nuclear binding energy:

E = Δmc²

where,

E = nuclear binding energy

Δm = mass defect

c = speed of light

Substitute the given value of mass defect and speed of light in above formula:

E = [tex]1.643 \times 10^{-28} kg\times(3 \times 10^{8})^{2}m^{2}/s^{2}[/tex]

E = [tex]14.787\times 10^{-12} kgm^{2}/s^{2}[/tex]

[tex]kgm^{2}/s^{2}[/tex] = Joule.

E= [tex]1.48\times 10^{-11} J[/tex]

Thus, nuclear binding energy of an atom is [tex]1.48\times 10^{-11} J[/tex].

Answer:

C. 1.48x10–11 J

Explanation:

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