Most Important Selected Qs for JEE AdvancedPhysicsNuclear Physics
Paragraph: Few atomic masses are given ₉₂²³⁸ U =238.05079 u , ₂^4 He =4.00260 u , ₉₀²³⁴ Th =234.04363 u ₁^1 H =1.007834, ₉₁²³⁷ ~Pa =237.065121 u . Answer the following questions on the basis of above data. Question: Calculate the energy released during the -decay of ₉₂²³⁸ U .
Options
- AQ =4.25 MeV
- BQ =8.5 MeV
- CQ =3.25 MeV
- DNone of these
Correct answer
A. Q =4.25 MeV
Step-by-step solution
₉₂²³⁸ U ₉₀²³⁴ Th + ₂^4 He . Here m =(238.05079-4.00260-234.04363) u E = mc ^2=4.24764 MeV