Saved Bookmarks
| 1. |
The isotopic masses of `._(1)^(2)H` and `._(2)^(4)He` are `2.0141` and `4.0026` amu respectively and the velocity of light in vacuum is `2.998xx10^(8) m//s`. Calculate the quantity of energy (in `J`) liberated when two mole of `._(1)^(2)H` undergo fusion to form one mole of `._(2)^(4)He` |
|
Answer» Correct Answer - `2.3xx10^(12)J` `DeltaE=Deltamxxc^(2)` `._(1)H^(2)+``_(1)H^(2)rarr ` `._(2)He^(4)` Mass defect `=Deltam` `=` Mass of `2(._(1)H^(2))-` Mass of `He` `=2xx2.0141-4.0026` `=4.0282-0026` `=0.0256g` `=0.256xx10^(-3)kg` `c=` speed of light `=2.998xx10^(8)m s^(-1)` `DeltaE=Deltamxxc^(20` `=0.0256xx10^(-3)xx(2.998xx10^(8))^(2)` `=0.23xx10^(13)J=2.3xx10^(12)J` |
|