what is the nuclear binding energy of an atom that has a mass defect of 1.643×10⁻²⁸ kg? use e = mc²…

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.)\n1.83×10⁻⁴⁶ j\n4.93×10⁻²⁰ j\n1.48×10⁻¹¹ j\n5.48×10⁴⁵ j
Answer
Explanation:
Step1: Identify values
$m = 1.643\times10^{-28}\text{ kg}$, $c = 3.00\times10^{8}\text{ m/s}$
Step2: Substitute into formula
$E=mc^{2}=(1.643\times 10^{-28})\times(3.00\times 10^{8})^{2}$
Step3: Calculate $(3.00\times 10^{8})^{2}$
$(3.00\times 10^{8})^{2}=3.00^{2}\times(10^{8})^{2}=9.00\times10^{16}$
Step4: Multiply by mass
$E=(1.643\times 10^{-28})\times(9.00\times10^{16})=(1.643\times9.00)\times10^{-28 + 16}$ $E = 14.787\times10^{-12}=1.4787\times 10^{-11}\approx1.48\times 10^{-11}\text{ J}$
Answer:
$1.48\times 10^{-11}\text{ J}$ (corresponding to the third option)