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Scienists are working hard to develop inclear fusion reactor Nocies of heavy hydrogen, `_(1)^(2)H` , known as deuteron and denoted by `D`, can be thought of as a candidate for fusion rector . The `D-D` reaction is `_(1)^(2) H + _(1)^(2) H rarr _(2)^(1) He + n+ energy` in the core of fasion reactor a gas of heavy hydrogen of `_(1)^(2) H` nucles and electrons is know as plasma . The nuclei move randonity in the reactor to take place Unally , the temperature in the reactor core are too ligh and to natrual will can be used to confine the to pleama for a time l_(0) before the particles by away from the case if `n` is the denasity (number volume ) of determines , the product` nt_(0) `is called Lavson number in one of the criteria , a reactor is termed successful if Lawson number is greater then `5 xx 10^(14) s//cm^(2)` it may be helpfull to use the following botczmann constant `lambda = 8.6 xx 10^(-5)eV//k, (e^(2))/(4 pi s_(0)) = 1.44 xx 10^(-9) eVm` Assume that two deuteron nuclei in the core of fasion reactor at temperacture energy `T` are moving toward each other, each with kinectic energy `1.5 kT` , whenn the seperation between them is large enogh to leglect coulomb potential energy . Also neglate any interaction from other particle in the core . The minimum temperature `T` required for them to reach a separation of `4 xx 10^(-15) m ` is in the rangeA. `1.0 xx 10^(9) K lt T lt 2. 0 xx 10^(9) K`B. `2.0 xx 10^(9) K lt T lt 3.0 xx 10^(9) K`C. `3.0 xx 10^(9) K lt T lt 4.0 xx 10^(9) K`D. `4.0 xx 10^(9) K lt T lt 5.0 xx 10^(9) K` |
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Answer» Correct Answer - A Appling conservation of machanical energy we get Loss of kinetic energy of two deuteron nuclei `= ` Gain in their potential energy `2 xx 1.5kT = (1)/(4 pi s_(0)) (e xx e)/®` `rArr 2 xx 1.5 xx(8.6 xx 10^(-5)(eV)/(k)) xx T = ((1.44 xx 10^(-9) eVm ))/(4 xx 10^(-15) m)` `rArr T = (1.44 xx 10^(-9))/(2 xx 1.5 xx 8.6 xx 10^(-5) xx 4 xx 10^(-15)) = 1.4 xx 10^(9)k` |
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