The effective Q for the proton–proton cycle of Fig. 43-11 is 26.2 MeV. (a) Express this as energy per kilogram of hydrogen consumed. (b) The power of the Sun is $\mathbf{3}\mathbf{.}\mathbf{9}\mathbf{x}{\mathbf{10}}^{\mathbf{26}}\mathbf{W}$. If its energy derives from the proton–proton cycle, at what rate is it losing hydrogen? (c) At what rate is it losing mass? (d) Account for the difference in the results for (b) and (c). (e) The mass of the Sun is $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{x}{\mathbf{10}}^{\mathbf{30}}\mathbf{kg}$. If it loses mass at the constant rate calculated in (c), how long will it take to lose 0.10% of its mass?

Let $\mathit{m}$be the mass of the protons sample, according to given equation, for each helium atom, four protons must be consumed, so the number of atoms in $\mathit{m}$is given by,

$\mathbf{N}\mathbf{=}\frac{\mathbf{m}}{\mathbf{4}{\mathbf{M}}_{\mathbf{P}}}$

The expression for the energy release per event is given by,

$\begin{array}{l}\mathbf{E}\mathbf{=}\mathbf{N}\mathbf{Q}\\ \mathbf{E}\mathbf{=}\frac{\mathbf{m}\mathbf{Q}}{\mathbf{4}{\mathbf{M}}_{\mathbf{P}}}\\ \frac{\mathbf{E}}{\mathbf{m}}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{4}{\mathbf{M}}_{\mathbf{P}}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)\end{array}$

Here, $\mathbf{Q}$is total energy, and${\mathbf{M}}_{\mathbf{P}}$ is mass of the proton.

Substitute all the known values in equation (1).

$$\begin{gathered} \frac{E}{m}=\frac{\left(26.2MeV\right)\left(1.602\times {10}^{-13}J/MeV\right)}{4\left(1.67\times {10}^{-27}kg\right)} \\ =6.3\times {10}^{14}J/kg \end{gathered}$$

Therefore, the energy per kilogram of hydrogen consumed is $6.3x{10}^{14}J/\mathrm{kg}$.

The expression to calculate the rate of losing hydrogen is given by,

$\frac{dm}{dt}=\frac{P}{\left(E/m\right)}...\dots \left(2\right)$

Here,$P$ is power of the Sun.

Substitute all the known values in equation (2).

$$\begin{gathered} \frac{dm}{dt}=\frac{3.9\times {10}^{26}J/s}{6.3\times {10}^{14}J/kg} \\ =6.2\times {10}^{11}kg/s \end{gathered}$$

Therefore, the rate of losing hydrogen is $6.2\times {10}^{11}kg/s$.

From the Einstein relation,

$E=m{c}^2$

But it is known that $P=\frac{dE}{dt}$, then

$\frac{dm}{dt}=\frac{P}{c^2}.....\left(3\right)$

Substitute all the known values in equation (3).

$$\begin{gathered} \frac{dM}{dt}=\frac{3.9\times {10}^{26}W}{{\left(3\times {10}^{8}m/s\right)}^2} \\ =4.34\times {10}^{9}kg/s \end{gathered}$$

Therefore, the rate of losing mass is$4.34\times {10}^{9}kg/s$ .

From above parts, it can be observed that the rate of which the sun losing hydrogen is greater than the rate of which the sun losing the mass.

$\frac{dm}{dt}>\frac{dM}{dt}$

As the protons are consumed, their mass is mostly turned into helium, which remains in the Sun.

The expression to calculate the time that the sun takes to lose $0.10\%$of its mass is given by,

$t=\frac{0.0010M}{\frac{dM}{dt}}..\dots 4)$

Here, $M$is mass of Sun.

Substitute all the known values in equation (4).

$$\begin{gathered} t=\frac{0.0010\left(2\times {10}^{30}kg\right)}{4.34\times {10}^{9}kg/s} \\ =4.60\times {10}^{17}s \\ =\left(4.60\times {10}^{17}s\right)\left(\frac{1y}{12\times 30\times 24\times 3600s}\right) \\ \approx 1.5\times {10}^{10}y \end{gathered}$$

Therefore, the required time is $1.5\times {10}^{10}y$.