Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.3k views
in Physics by (87.2k points)
closed by
(a) Calculate the energy released by the fission of `2g` of `._(92)U^(235)` in `kWh`. Given that the energy released per fission is `200MeV`.
(b) Assuming that `200MeV` of enrgy is released per fission of uranium atom, find the number of fissions per second required to released `1` kilowatt power.
(c) Find the amount of energy produced in joules due to fission of `1g` of `._(92)U^(235)`assuming that `0.1%` of mass is transformed into enrgy.`._(92)U^(235) = 235 amu`, Avogadro number `= 6.023 xx 10^(23)`

1 Answer

0 votes
by (84.6k points)
selected by
 
Best answer
(a) Number of fissions `=` Number of atoms
`= (m)/(M)N_(A) = (2)/(235) xx 6.023 xx 10^(23)`
Total energy released
`=(2)/(235) xx 6.023 xx 10^(23) xx 200 xx 10^(6) xx 1.6 xx10^(-19)J`
`= 16.4 xx 10^(10) J`
`= (16.4xx10^(10))/(3.6xx10^(6)) = 4.55 xx 10^(4)kWh`
`(1kWh = 3.6 xx 10^(6)J)`
(b) `E = 1kW = 10^(3) J//sec`
Number of fissions/sec `= (10^(3))/(200xx1.6xx10^(13)) = 3.125xx10^(13)`
(c) Mass used `m = (0.1)/(100) xx 1=10^(-3) g`
Number of fissions `= (10^(-3))/(235) xx 6.023 xx 10^(23)`
Enegry produced `= (6.023xx10^(20))/(235) xx 200 xx 1.6 xx 10^(-13)`
`= 8.2 xx 10^(7) J`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...