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 $$\left( {^4He} \right).$$ At what rate $$\frac{{dm}}{{dt}}$$ 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.
$$1.6 \times {10^{10}}kg/s$$
B.
$$2.3 \times {10^9}kg/s$$
C.
$$6.2 \times {10^{11}}kg/s$$
D.
$$5.5 \times {10^{10}}kg/s$$
Answer :
$$6.2 \times {10^{11}}kg/s$$
Solution :
The rate $$\frac{{dm}}{{dt}}$$ can be calculate as;
Power, $$P = \frac{{dE}}{{dt}} = \frac{{dE}}{{dm}} \times \frac{{dm}}{{dt}} = \frac{{\Delta E}}{{\Delta m}} \times \frac{{dm}}{{dt}}$$
$$\therefore \frac{{dm}}{{dt}} = \frac{{\Delta m}}{{\Delta E}}P\,.......\left( {\text{i}} \right)$$
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
$$\eqalign{
& \frac{{dm}}{{dt}} = \frac{{\Delta m}}{{\Delta E}}P = \frac{{4\left( {1.67 \times {{10}^{ - 27}}} \right)}}{{4.20 \times {{10}^{ - 12}}}} \times \left( {3.90 \times {{10}^{26}}} \right) \cr
& = 6.2 \times {10^{11}}kg/s \cr} $$