Question 258: Atoms & Nuclei

Question

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]

Accepted Answer

$${}_3^7Li$$ + $${}_1^1H$$ $$ 	o $$ 2($${}_2^4He$$)
$$\Delta $$m = $$\left[ {{M_{Li}} + {M_H}} ight] - 2\left[ {{M_{He}}} ight]$$
= (7.0160 + 1.0079) - 2 $$ 	imes $$ 4.0003
= 0.0187
Energy released in 1 reaction = $$\Delta $$mc
2
In use of 7.016 u Li energy is $$\Delta $$mc
2
.
In use of 1 gm Li energy is $${{\Delta m{c^2}} \over {{m_{Li}}}}$$.
In use of 20 gm energy is = $${{\Delta m{c^2}} \over {{m_{Li}}}} 	imes 20$$
= $${{0.087 	imes 1.6 	imes {{10}^{ - 19}} 	imes {{10}^9}} \over {7.016 	imes 1.6 	imes {{10}^{ - 24}}}} 	imes 20$$ J
= 0.05 $$ 	imes $$ 10
14
J
= 1.33 $$ 	imes $$ 10
6
kWh

Answer

6.82 $$ 	imes $$ 105

Answer

4.5 $$ 	imes $$ 105

Answer

8 $$ 	imes $$ 106

Answer

1.33 $$ 	imes $$ 106

Atoms & Nuclei

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]

Choose the correct answer: