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The following fusion reaction take place...

The following fusion reaction take place `2_1^2Ararr_2^3B+n+3.27` MeV. If 2 kg of `._1^2A` is subjected to the above reaction, the energy released is used to light a 100 W light a lamp, how long will the lamp glow ?

A

`2 xx 10^(6)` years

B

`3 xx 10^(5)` years

C

`5 xx 10^(4)` years

D

`7 xx 10^(3)` years

Text Solution

Verified by Experts

The correct Answer is:
C
No. of atom in 2g of deuterium `=6.023 xx 10^(23)`
No. of atom in 2g of deuterium `= (6.023 xx 10^(23))/(2) xx 2000= 6.023 xx 10^(26)`
Given, energy released in fusion of two deuterium = 3.27 MeV
`therefore` Total energy `= (3.27)/(2) xx 6.023 xx 10^(26) MeV`
`=15.75 xx 10^(13)J`
Given, Power (P) = 100W = 100 J / sec
`t= (E)/(P) = (15.75 xx 10^(13))/(100) = 15.75 xx 10^(11)` sec
`rArr t` (in years) `= (15.75 xx 10^(11))/(60 xx 60 xx 24 xx 365)= 5 xx 10^(4)` year
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