You are given that Li 3 7 = 7 . 0160 u , Mass of Mass of He 2 4 = 4 . 0026 u and Mass of He 1 1 = 1 . 0079 H…

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]
  1. 4.5×105
  2. 8×106
  3. 6.82×105
  4. 1.33×106

Solution

E=ΔmC2

E=(1.0079+7.0160-2(4.0026)×931

=1.33×106

Asked in: JEE Main 2020 (06 Sep Shift 1)

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