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

A star initially has `10^(40) ` deuterons it product 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 deuteron supply of the average power radiated by the state is `10^(16) W` , the deuteron supply of the state 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`.

A

`10^(6) s.`

B

`10^(8) s.`

C

`10^(12) s.`

D

`10^(16) s`.

Text Solution

Verified by Experts

The correct Answer is:
C
`_(1)H^(2) + _(1)H^(2) rarr _(1)H^(3) + p`
`_(1)H^(2) + _(1)H^(3) rarr _(2)He^(4) +n`
`Net reaction 3_(1)H^(2) rarr _(2)He^(4) + p + n`
`deltam = 3 (2.014) - [4.001 + 1.007 + 1.008 ] = 0.026 `
`3deuterons release 3.87 xx 10^(-12)J`
`:. 10^(40) deuterons release = (3.87 xx 10^(-12) xx 10^(40))/(3)`
` = 1.29 xx 10^(28) J`
`p = (E)/(t) rArr t = (E)/(p) = (1.29 xx 10^(28))/(10^(16)) = 1.29 xx 10^(12) sec`
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