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A star initially has 10^40 deuterons. It...

A star initially has `10^40` deuterons. It produces energy via the processes `_1^2H+_1^2Hrarr_1^3H+p` and `_1^2H+_1^3Hrarr_2^4He+n`. Where the masses of the nuclei are
`m(`^2H)=2.014` amu, `m(p)=1.007` amu, `m(n)=1.008` amu and `m(`^4He)=4.001` amu. If the average power radiated by the star is `10^16 W`, the deuteron supply of the star is exhausted in a time of the order of

A

(a) `10^6s`

B

(b) `10^8s`

C

(c) `10^12s`

D

(d) `10^16s`

Text Solution

Verified by Experts

The correct Answer is:
C
Combining two given equations,
we have, `3_1H^2=2He^4+p+n`
`Deltam=3xx2.014-4.001-1.007-1.008`
`=0.026u`
Energy released by 3 deuterons
`=0.026xx93.15xx1.6xx10^-13J`
`=3.9xx10^-12J`
Now, `(10^16xxt)=((10^40)/(3))(3.9xx10^-12)`
Solving we get,
`t=1.3xx10^12s`
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