Question

Deuterium (2H), when heated to sufficiently high tempera- ture, undergoes a nuclear fusion reaction that results in the production of helium. The reaction proceeds rapidly at a temperature, T, at which the average kinetic energy of the deuterium atoms is 8 × 10216 J. (At this temperature, deu- terium molecules dissociate completely into deuterium atoms.)
(a) Calculate T in kelvins (atomic mass of 2H 5 2.015).
(b) For the fusion reaction to occur with ordinary H atoms, the average energy of the atoms must be about 32 × 10216 J. By what factor does the average speed of the 1H atoms differ from that of the 2H atoms of part (a)?

Answer 1

Answer:

Temperature=3.86*10^7K

V1=$V_{2}$1.4

Explanation:

Deuterium (2H), when heated to sufficiently high tempera- ture, undergoes a nuclear fusion reaction that results in the production of helium. The reaction proceeds rapidly at a temperature, T, at which the average kinetic energy of the deuterium atoms is 8 × 10216 J. (At this temperature, deu- terium molecules dissociate completely into deuterium atoms.)

(a) Calculate T in kelvins (atomic mass of 2H 5 2.015).

(b) For the fusion reaction to occur with ordinary H atoms, the average energy of the atoms must be about 32 × 10216 J. By what factor does the average speed of the 1H atoms differ from that of the 2H atoms of part (a)?

k.E =3/2kT

K.E is the kinetic energy

k=boltzmann constant

T=Temperature in kelvin

juxtaposing the given values we have

T=2(KE)/3k

T=2*8*10^-16/(3*1.38*10^-23)

T=3.86*10^7K

Temperature=3.86*10^7K

b. average speed V=$\sqrt{8RT/\pi }M$

R is the gas constant

T absolute temperature in kelvin

M molar mass

since R.T,PI are constant

we are comparing between hydrogen and helium particles

V=$\alpha \sqrt{1/M}$

V1/V2=$\sqrt{M2/M1}$

V1=V2$\sqrt{2/1}$

V1=V2* 1.4

V1=$V_{2}$1.4