Question

The sun radiates energy at the rate of $4\times {10}^{26}Joule{\sec }^{-1}$. If the energy of fusion process$4_1^{1}H{\to }_2^{4}He+2_1^{0}e$ is $27MeV$, calculate amount of hydrogen atoms that would be consumed per day for the given process.

Answer 1

$27MeV=27\times {10}^6\times 1.6\times {10}^{-19}=43.2\times {10}^{-13}J$

Energy radiated by the sun per day $=4\times {10}^{26}\times 3600\times 24J{day}^{-1}=34.56\times {10}^{30}J{day}^{-1}$

$43.2\times {10}^{-13}J$ of energy is obtained from $=4amu$ of $H=4\times 1.66\times {10}^{-24}g$ of $H$

$34.56\times {10}^{30}J$ of energy is obtained from $=\frac{4\times 1.66\times {10}^{-24}}{43.2\times {10}^{-13}}\times 34.56\times {10}^{30}=5.31\times {10}^{19}g$