E=ΔmC2
E=(1.0079+7.0160−2(4.0026)×931
=1.33×106
JEE Main 2020 — Physics Modern Physics
You are given that Li37=7.0160u,Mass of Mass of He24=4.0026u and Mass of He11=1.0079H When 20g of Li37 is converted into 24 He by proton capture, the energy liberated, (in kWh ), is : [Mass of nucleon =1GeV/c2]
Held on 6 Sept 2020 · Verified 6 Jul 2026.
4.5×105
8×106
6.82×105
1.33×106
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