JEE PYQ: Nuclear Physics Question 16
Question 16 - 2020 (06 Sep Shift 1)
You are given that mass of $^7_3\text{Li} = 7.0160$ u, Mass of $^4_2\text{He} = 4.0026$ u and Mass of $^1_1\text{H} = 1.0079$ u.
When 20 g of $^7_3\text{Li}$ is converted into $^4_2\text{He}$ by proton capture, the energy liberated, (in kWh), is: [Mass of nucleon = 1 GeV/c$^2$]
(1) $4.5 \times 10^5$
(2) $8 \times 10^6$
(3) $6.82 \times 10^5$
(4) $1.33 \times 10^6$
Show Answer
Answer: (4)
Solution
$^7_3\text{Li} + ^1_1\text{H} \longrightarrow 2(^4_2\text{He})$ $\Delta m \to [m_{\text{Li}} + m_H] - 2[M_{\text{He}}]$ Energy released $= \Delta mc^2$ In use of 1 g Li energy released $= \frac{\Delta mc^2}{m_{\text{Li}}}$ In use of 20 g energy released $= \frac{\Delta mc^2}{m_{\text{Li}}} \times 20$ g $= \frac{[(7.016 + 1.0079) - 2 \times 4.0026]u \times c^2}{7.016 \times 1.6 \times 10^{-24}} \times 20$ $= \left(\frac{0.0187 \times 1.6 \times 10^{-19} \times 10^9}{7.016 \times 1.6 \times 10^{-24}} \times 20\right) = 480 \times 10^{10}$ J $\because 1$ J $= 2.778 \times 10^{-7}$ kWh $\therefore$ Energy released $= 480 \times 10^{10} \times 2.778 \times 10^{-7}$ $= 1.33 \times 10^6$ kWh
BETA
