You are given that Mass of ${ }_{3}^{7} \mathrm{Li}=7.0160 \mathrm{u}$, Mass of ${ }_{2}^{4} \mathrm{He}=4.0026 \mathrm{u}$ and ${ }_{1}^{1} \mathrm{H}=1.0079 \mathrm{u}$. When 20 g of ${ }_{3}^{7} \mathrm{Li}$ is converted into ${ }_{2}^{4} \mathrm{He}$ by proton capture, the energy liberated, (in kWh), is [Mass of nucleon $=\frac{1 \mathrm{GeV}}{\mathrm{c}^{2}}$]

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Modern Physics

You are given that Mass of

_{3}^{7} Li = 7.0160 u

, Mass of

_{2}^{4} He = 4.0026 u

and

_{1}^{1} H = 1.0079 u

. When 20 g of

_{3}^{7} Li

is converted into

_{2}^{4} He

by proton capture, the energy liberated, (in kWh), is [Mass of nucleon

= \frac{1 GeV}{c ^{2}}

]

Answer: (D)

Each proton-capture event is

X_{3}^{7} X_{23}^{27} Li + X_{1}^{1} X_{21}^{21} H 2 X_{2}^{4} X_{22}^{24} He

, and the mass lost in it is

Δ m = 7.0160 + 1.0079 - 2 (4.0026) = 0.0187 u

. Converting with

1 u \equiv 931.5 MeV

, the energy released per event is

Q = 0.0187 \times 931.5 \approx 17.4 MeV

. In

20 g

of lithium there are

N = \frac{20}{7} \times 6.02 \times 1 0^{23} \approx 1.72 \times 1 0^{24}

atoms, each giving one such event, so the total energy is

E = N Q \approx 1.72 \times 1 0^{24} \times 17.4 \times 1.6 \times 1 0^{- 13} J \approx 4.8 \times 1 0^{12} J

. Dividing by

3.6 \times 1 0^{6} J

per kWh gives

E \approx 1.33 \times 1 0^{6} kWh

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Single correct option