JEE MAIN - Physics (2020 - 6th September Morning Slot - No. 13)
You are given that Mass of $${}_3^7Li$$ = 7.0160u,
Mass of $${}_2^4He$$ = 4.0026u
and Mass of $${}_1^1H$$ = 1.0079u.
When 20 g of $${}_3^7Li$$ is converted into $${}_2^4He$$ by proton capture, the energy liberated, (in kWh), is :
[Mass of nucleon = 1 GeV/c2]
Mass of $${}_2^4He$$ = 4.0026u
and Mass of $${}_1^1H$$ = 1.0079u.
When 20 g of $${}_3^7Li$$ is converted into $${}_2^4He$$ by proton capture, the energy liberated, (in kWh), is :
[Mass of nucleon = 1 GeV/c2]
6.82 $$ \times $$ 105
4.5 $$ \times $$ 105
8 $$ \times $$ 106
1.33 $$ \times $$ 106
Explanation
$${}_3^7Li$$ + $${}_1^1H$$ $$ \to $$ 2($${}_2^4He$$)
$$\Delta $$m = $$\left[ {{M_{Li}} + {M_H}} \right] - 2\left[ {{M_{He}}} \right]$$
= (7.0160 + 1.0079) - 2 $$ \times $$ 4.0003
= 0.0187
Energy released in 1 reaction = $$\Delta $$mc2
In use of 7.016 u Li energy is $$\Delta $$mc2.
In use of 1 gm Li energy is $${{\Delta m{c^2}} \over {{m_{Li}}}}$$.
In use of 20 gm energy is = $${{\Delta m{c^2}} \over {{m_{Li}}}} \times 20$$
= $${{0.087 \times 1.6 \times {{10}^{ - 19}} \times {{10}^9}} \over {7.016 \times 1.6 \times {{10}^{ - 24}}}} \times 20$$ J
= 0.05 $$ \times $$ 1014 J
= 1.33 $$ \times $$ 106 kWh
$$\Delta $$m = $$\left[ {{M_{Li}} + {M_H}} \right] - 2\left[ {{M_{He}}} \right]$$
= (7.0160 + 1.0079) - 2 $$ \times $$ 4.0003
= 0.0187
Energy released in 1 reaction = $$\Delta $$mc2
In use of 7.016 u Li energy is $$\Delta $$mc2.
In use of 1 gm Li energy is $${{\Delta m{c^2}} \over {{m_{Li}}}}$$.
In use of 20 gm energy is = $${{\Delta m{c^2}} \over {{m_{Li}}}} \times 20$$
= $${{0.087 \times 1.6 \times {{10}^{ - 19}} \times {{10}^9}} \over {7.016 \times 1.6 \times {{10}^{ - 24}}}} \times 20$$ J
= 0.05 $$ \times $$ 1014 J
= 1.33 $$ \times $$ 106 kWh
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