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Blazing goIITian

Joined:	8 Dec 2006
Post:	351

27 Mar 2007 18:36:00 IST

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625

objective question

Engineering Entrance , Medical Entrance , AIPMT , JEE Main , AIIMS , JEE Advanced , Physics , Modern Physics

consider the radius of deuterium and tritium to be approximately 1.5*10^-15m each and that for the deuterium-tritium fusion reaction,the mass defect is approximately 18 MeV.Then,

A) kinetic energy needed to overcome coulomb repulsion is approximately 8*10^-14 J.

B) the temperature to which the gases must be heated to initiate reaction is of the order of 10^9 K.

C)the amount of energy liberated in the reaction is approximately 18 MeV.

D) mass defect involved in the reaction is 0.0123 amu.

give the correct answer along with proper explanation and if u think that suppose options b and c are correct and a is not correct,then plz tell why a is not corrrect..expecting a good and early reply..

Comments (2)

nishant dash

Blazing goIITian

Joined: 8 Dec 2006

Posts: 351

29 Mar 2007 11:01:34 IST

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no one good enough to answer my question here?

CyBorG

Forum Expert

Joined: 6 Jan 2007

Posts: 706

7 Apr 2007 13:58:43 IST

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Kinetic energy needed to overcome coulomb repulsion KE =(q1)(q2)/(4piE0r)
Here r=1.5*10^-15*2=3*10^-15m
q1=e
q2=e
Also KE=3kT/2
So T=2KE/3k where k is boltzmann constant.
So T is of the order 10^9
Also energy liberated should be 18Mev
So the answers are B) and C)
Mass defect=18/931=0.01933 amu

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