Question

The proton -proton mechanism that accounts for energy production in the sun releases $26.7\, MeV$ energy for each event. In this process, protons fuse to form an alpha particle $\left({ }^4 He \right)$. At what rate $\frac{d m}{d t}$ is hydrogen being consumed in the core of the sun by the p-p cycle? Power of sun is $3.90 \times 10^{26} W$.

• A. $1.1 \times 10^{11} kg / s$
• B. $2.3 \times 10^{10} kg / s$
• C. $3.2 \times 10^9 kg / s$
• D. $6.2 \times 10^{11} kg / s$

Answer 1

Answer: (D)

The rate $d m / d t$ can be calculate as ;
Power, $P=\frac{d E}{d t}=\frac{d E}{d m} \times \frac{d m}{d t}=\frac{\Delta E}{\Delta m} \times \frac{d m}{d t}$
$\therefore \frac{d m}{d t}=\frac{\Delta m}{\Delta E} P$....(i)
We known that $26.2 \,MeV =4.20 \times 10^{-12} J$ of thermal energy is produced when four protons are consumed.
This is $\Delta E =4.20 \times 10^{-12} J$ for $\Delta m =4 \times\left(1.67 \times 10^{-27} kg \right)$.
Substituting these values in equation (i), we have
$\frac{d m}{d t}=\frac{\Delta m}{\Delta E} P=\frac{4\left(1.67 \times 10^{-27}\right)}{4.20 \times 10^{-12}} \times\left(3.90 \times 10^{26}\right) $
$=6.2 \times 10^{11} kg / s$