Question

calculate the energy released when 2amu of mass is converted into energy.(1 amu=1.66×10^-27kg)​

Answer 1

from Enstien's relativity theory,

E = mC²

where ,E is the amount of energy, m is mass of particle and C is velocity of light

given, mass of particle, m = 2 a.m.u

= 2 × 1.66 × 10^-27 Kg

= 3.32 × 10^-27 Kg

velocity of light in vaccum, C = 3 × 10^8 m/s

so, energy = 3.32 × 10^-27 × (3 × 10^8)²

= 3.32 × 10^-27 × 9 × 10^16 J

= 29.88 × 10^-11 J

= 2.988 × 10^-10 J

hence, 2.988 × 10^-10 J energy released when 2 a.m.u of mass is converted into energy.

Answer 2

Answer:

We know, E = mC²

where ,E is the amount of energy, m is mass of particle and C is velocity of light given, mass of particle, m = 2 a.m.u

= 2 × 1.66 × 10^-27 Kg = 3.32 × 10^-27 Kg

Velocity of light in vacuum, C = 3 × 10^8 m/s

Therefore, energy = 3.32 × 10^-27 × (3 × 10^8)²

= 3.32 × 10^-27 × 9 × 10^16 J = 29.88 × 10^-11 J = 2.988 × 10^-10 J

So, 2.988 × 10^-10 J energy released when 2 a.m.u of mass is converted into energy.