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 $(^{4} H e)$ . The rate at which hydrogen being consumed in the core of the sun by the p-p cycleis $\frac{d m}{d t}$ is $x \times 10^{11} k g / s$ . Power of sun is $3.90 \times 10^{26} W$ . Find the value of x.
[Round-off to two decimal places]

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answer is 6.21.

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Detailed Solution

The rate dm/dt can be calculate as;

Power, $P = \frac{d E}{d t}$
$= \frac{d E}{d m} \times \frac{d m}{d t} = \frac{Δ E}{Δ m} \times \frac{d m}{d t}$
$∴ \frac{d m}{d t} = \frac{Δ m}{Δ E} P ..... (i)$
We known that 26.2 MeV $= 4.192 \times 10^{- 12} J$ of thermal energy is produced when four protons are consumed.

This is $Δ E = 4.20 \times 10^{- 12} J$ for $Δ m = 4 \times (1.67 \times 10^{- 27} k g) .$
Substituting these values in equation (i), we have $\frac{d m}{d t} = \frac{Δ m}{Δ E} P$
$= \frac{4 (1.67 \times 10^{- 27})}{4.192 \times 10^{- 12}} \times (3.90 \times 10^{26})$
$= 6.21 \times 10^{11} k g / s$