what is the nuclear binding energy of an atom that has a mass defect of 5.0446 × 10⁻²⁹ kg? use e = mc²…

what is the nuclear binding energy of an atom that has a mass defect of 5.0446 × 10⁻²⁹ kg? use e = mc². (remember: the speed of light is approximately 3.00 × 10⁸ m/s.)\no 5.61 × 10⁻⁴⁶ j\no 1.51 × 10⁻²⁰ j\no 4.54 × 10⁻¹² j\no 1.78 × 10⁴⁵ j

what is the nuclear binding energy of an atom that has a mass defect of 5.0446 × 10⁻²⁹ kg? use e = mc². (remember: the speed of light is approximately 3.00 × 10⁸ m/s.)\no 5.61 × 10⁻⁴⁶ j\no 1.51 × 10⁻²⁰ j\no 4.54 × 10⁻¹² j\no 1.78 × 10⁴⁵ j

Answer

Explanation:

Step1: Identify the values

$m = 5.0446\times10^{-29}\text{ kg}$, $c = 3.00\times 10^{8}\text{ m/s}$

Step2: Substitute into the formula

$E=mc^{2}=(5.0446\times 10^{-29})\times(3.00\times 10^{8})^{2}$ First calculate $(3.00\times 10^{8})^{2}=3.00^{2}\times(10^{8})^{2}=9.00\times 10^{16}$ Then $E=(5.0446\times 10^{-29})\times(9.00\times 10^{16})$ Using the rule of exponents $a^{m}\times a^{n}=a^{m + n}$, we have $E=(5.0446\times9.00)\times10^{-29 + 16}=45.4014\times10^{-13}=4.54014\times 10^{-12}\text{ J}\approx4.54\times 10^{-12}\text{ J}$

Answer:

C. $4.54\times 10^{-12}\text{ J}$