A star initially has 10^(40) deuterons....

A star initially has `10^(40) ` deuterons. It produces energy via the process `_(1)H^(2) + _(1)H^(2) + rarr _(1) H^(3) + p. `and `_(1)H^(2) + _(1)H^(3) + rarr _(2) He^(4) + n` .If the average power radiated by the state is `10^(16) W` , the deuteron supply of the star is exhausted in a time of the order of .
The masses of the nuclei are as follows:
`M(H^(2)) = 2.014 amu,`
`M(p) = 1.007 amu, M(n) = 1.008 amu, M(He^(4)) = 4.001 amu`.

`10^(6) s`

`10^(8) s`

`10^(12) s`

`10^(16) s`

Verified by Experts

The correct Answer is:

Add equation `3._(1)^(2)H rarr ._(2)^(4)HE + p + n`
(A) `Q = [-[m(._(2)^(4)He)+m(p) + m(n)] + 3m(._(1)^(2)H)] xx 931 MeV`
(B) `10^(16)W = (Q)/(t)`

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The reaction: 2MnO_(4)^(-) + 3Mn^(2+) + 2H_(2)O rarr 5MnO_(2) + 4H^(+) is catalysed by…..

`NiSO_(4)`

`ZnSO_(4)`

`Mn^(2+)`

`Fe^(3+)`

`Delta H_(f) = - Delta H_(1) + (Delta H_(2) )/( 2) - (3)/(2) Delta H_(3)`

`Delta H_(f) = - Delta H_(2) + (Delta H_(2) )/( 2) - (3)/(2) Delta H_(3)`

`Delta H_(f) = - Delta H_(2) + (-Delta H_(2) )/( 2) - (3)/(2) Delta H_(3)`

none of above