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The isotopic masses of .(1)H^(2) and .(2...

The isotopic masses of `._(1)H^(2)` and `._(2)He^(4)` are 2.0141 and 4.0026 amu, respectively. Calculate the quantity of energy liberated when two moles of `._(1)H^(2)` undergo fission to form 1 mol of `._(2)He^(4)`. The velocity of light in vacuum is `3.0 xx 10^(8) m s^(-1)`.

Text Solution

Verified by Experts

The fusion reaction is
`._(1)H^(2) + ._(1)H^(2) rarr ._(2)H^(4) + "Energy"`
Mass defectsd `= 2 xx` mass of `._(1)H^(2)`- Mass of `._(2)H^(4)`
`= 2 xx 2.0141 - 4.0026 = 0.0256 "amu"`
Energy released for fusion of two atoms
`= Delta mc^(2)`
`= 0.0256 xx 1.66 xx 10^(-27) xx (3.0 xx 10^(8))^(2)`
`= 3.8 xx 10^(-12) J` `[1 "amu" = 1.66 xx 10^(-27) kg]`
Energy released per mol of `._(2)He^(4)`
`= 3.8 xx 10^(-12) xx 6.23 xx 10^(23)`
`= 2.3 xx 10^(12)J`
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