wiz-icon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

It is proposed to use the nuclear fusion reaction 21H+21H42He in a nuclear reactor with an electrical power rating of 200 MW. If the energy from the above reaction is used with 25 percent efficiency in the reactor, how many grams of deuterium fuel will be needed per day? (The masses of 21H and 42He are 2.0141 amu and 4.0026 amu respectively.)

A
290 gm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
100 gm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
130 gm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
230 gm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 130 gm
Mass defect occuring in one fusion reaction,
Δm=(2×2.01414.0026) amu=0.256 amu

Energy released =0.256×931 MeV

Energy used in reactor per fusion reaction =25100×0.0256×931 MeV
=5.9584 MeV=9.5334×1013 J

Total energy required per day =(200 MW)×(24×60×60) s

Mass of deuterium fuel needed per reaction =2×2.0141 amu=4.02826.023×1023 g=0.6691×1023 g

Masses of deuterium required = 0.6691×1023×200×106×24×60×609.5334×1013=130 g

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Nuclear Fusion
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon