Question

50 points! The isotope shown has a mass of 14.003241 amu. Calculate how much energy is released from the binding of 2.530x10 moles of this isotope. All particles in the nucleus are shown.​

Answer 1

Answer : The amount of energy released is,
`3.189* 10^(16)J`

Explanation :

Formula used to calculate the amount of energy released is:

`E=mc^2`

where,

E = amount of energy released

m = mass of isotope

c = speed of light =
`3* 10^8m/s`

First we have to calculate the mass of isotope.

Mass of isotope = Moles of isotope × Atomic mass of isotope

Mass of isotope = (2.530 × 10)mol × 14.003241 amu

Mass of isotope = 354.3 g = 0.3543 kg

Now we have to calculate the amount of energy released.

`E=mc^2`

`E=(0.3543kg)* (3* 10^8m/s)^2`

`E=3.189* 10^(16)J`

Thus, the amount of energy released is,
`3.189* 10^(16)J`

Answer 2

E = 3.1885 E16 J

Step-by-step explanation:

• E = mc²

∴ c = 3 E8 m/s

∴ m = (2.53 E1)(14.003241 amu) = 354.282 g isotope

⇒ E = (354.282 g)(3 E8)²

⇒ E = 3.1885 E19 g.m²/s²

⇒ E = 3.1885 E16 Kg.m²/s² = 3.1885 E16 J