Question

The proton -proton mechanism that accounts for energy production in the sun releases $26.7MeV$ energy for each event. In this process, protons fuse to form an alpha particle $\left({}^{4}He\right)$. At what rate $\frac{dm}{dt}$ 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 $dm/dt$ can be calculate as ;
Power, $P=\frac{dE}{dt}=\frac{dE}{dm}\times \frac{dm}{dt}=\frac{\Delta E}{\Delta m}\times \frac{dm}{dt}$
$\therefore \frac{dm}{dt}=\frac{\Delta m}{\Delta E}P$....(i)
We known that $26.2MeV=4.20\times 10^{-12}J$ 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
$\frac{dm}{dt}=\frac{\Delta m}{\Delta E}P=\frac{4\left(1.67\times 10^{-27}\right)}{4.20\times 10^{-12}}\times \left(3.90\times 10^{26}\right)$
$=6.2\times 10^{11}kg/s$