Principles of Nuclear Technology Course
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Q1 . Activity of one gm of the Shroud of Turin i .e . R is calculated below R (3 counts per min 100 300 counts per min 5 counts /s 5 Bq Activity of one gm of the Shroud of Turin i .e . Ro is calculated below Ro 512 pCi (512 10-12 3 .7 1010 Bq 18 .94 Bq R and Ro is related by the following equation ?t Where ? is decay constant of C14 and given by 0 .693 /t1 /2 t1 /2 of C14 is 5760 years and t is life of the Shroud of Turin Therefore , life of the Shroud of Turin will be The calculation shows that the sample is 11072 yrs old . This simply means that the shroud is not real .Q2 . Let us assume there were N atoms of Pu238 in the RHU after 8 years .Then , Rate of disintegration will be R N ?Where ? is the decay constant of Pu238 and is given as (0 .693 /t1 /2 where t1 /2 of Pu238 is 87 .7 years 87 .7 365 24 3600 s 2 .77 109 s Now , disintegration of each Pu238 atom gives a ? particle having energy E 5 .45 MeV 5 .45 1 .6 10-13 8 .72 10-13 which is absorbed in the RHU .Therefore , power due to disintegration of Pu238 will be For this power to be 1 .5W 1 .5 Now let us try to calculate number of Pu238 at the time of launch let us take that as No .No and N is related by the following equation :h j ‘- J L Z \ ?A TZ ‘r tz jf Noe- ?t 7 .33 1021 Therefore , the initial mass of Pu238 will be Q3 . Let Do be the dose rate without the door and D is the dose rate after the Pb door of thickness t has been installed . Then , thickness t has to be calculated such that D and Do is related by the following equation D Doe- ?t Where ? is the linear attenuation coefficient of Pb for 2 MeV ?-rays and t is thickness of Pb door .Linear attenuation coefficient of Pb for 2 MeV ?-rays is 51 .8 m-1 Therefore , thickness of the Pb door for 100 times attenuation of the 2 MeV ? dose rate will be Area of the door A 8 sq ft 8 30 .48 30 .48 cm2 7432 .2432 cm2 Volume of the door A t 7432 .2432 8 .892 cm3 66087 .51 cm3 Mass of the lead door will be m ?V 11 .35 66087 .51 gm (11 .35 66087 .51 /453 .6 lb 1653 .65 lb Therefore cost of the lead door will be C 0 .80 1653 .65 1323 Reference :Glassstone S and Sesonske A . “Nuclear Reactor Engineering ‘ Volume 1 .Chapman Hall Inc . New York…

