Question

A star has 10$^{40}$ deuterons. It produces energy via the process
$_1$H$^{2}$ + $_1$H$^{2}$ $\rightarrow$ $_1$H$^{3}$ + $_1$H$^{1}$
$_1$H$^{2}$ + $_1$H$^{3}$ $\rightarrow$ $_2$He$^{4}$ + $_0$n$^{1}$
If the average power radiated by the star is 10$^{16}$ W, then the deuteron supply of the star is exhausted in a time of the order of:
(A) 10$^{6}$ s
(B) 10$^{8}$ s
(C) 10$^{12}$ s
(D) 10$^{16}$ s

Answer 1

Hint:It is a numerical question based on the concept of mass defect, and the energy released. With the help of calculation of mass defect, and the energy released; the average power is given; the time required can be calculated. We have the formula of mass defect, i.e.
$\Delta$m = mass of products – mass of reactants. We can use this formula for mass defect.

Complete step by step solution:
> Now, we know that we are given two reactions. In the first reaction, there is formation of a proton, and in the second reaction, there is formation of a helium nucleus, and a neutron.
Thus, by adding these two equations we get,
3($_1$H$^{2}$) $\rightarrow$ $_2$He$^{4}$ + $_0$n$^{1}$ + $_1$H$^{1}$
> Now, the mass of the helium nucleus is 4.014, the deuterium is 2.014, and the neutron has the mass 1.007, and the proton has the mass 1.0008.
- Thus, by using the mass defect, or mass energy conservation; as mentioned i.e. mass defect is equal to mass of the reactants (left-side) subtracted from the mass of products (right-side).
- Therefore, it can be written as
$\Delta$ m = (3 $\times$ 2.014 – 4.001 – 1.007 – 1.0008) = 0.026 amu
> Now, we can calculate the energy released in this equation, i.e.
- Energy released value = 0.026 $\times$ 931 MeV, (1 MeV = 1.6 $\times$ 10$^{-13}$J)
Thus, energy released = 3.87$\times$ 10$^{-12}$ J
- The value of energy released is the consumption of energy by 3 deuteron atoms.
> Now, we have to calculate for the 10$^{40}$ deuterons
Energy released = $\dfrac{10^{40} \times 3.87 \times 10^{-12}}{3}$= 1.29 $\times$ 10$^{28}$ J
- In the question, the average power radiated is given i.e. 10$^{16}$ W, is equal to 10$^{16}$ J/s.
> Now, further let us calculate the time required to exhaust the deuteron supply of star.
We can calculate it by dividing the energy released to that of the average power radiated, i.e. $\dfrac{1.29 \times 10^{28}}{10^{16}}$ = 1.29 $\times$ 10$^{12}$ s =10$^{12}$ s (approx.)
In the last, we can conclude that the deuteron supply of star is exhausted in the time order of 10$^{12}$ Hence, the correct option is (C).

Note: We can calculate the energy consumption of 3 deuteron atoms by mass-energy conservation, i.e. by multiplying the energy released in the formula of mass defect only. There is no need of confusion within the two formulae.