QJEEM20Ph6198: JEE Main 2020 Physics - Class 12 JEEM20Ph6198 Solution: You are given that Mass of = 7.0160u, Mass of = 4.0026u and…

Question

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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

\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

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