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Standard VII

Chemistry

Sub-Atomic Particles

A star has ...

Question

A star has $10^{40}$ deuterons. It produces energy via the process
$_{1} H^{2} +_{1} H^{2} \to_{1} H^{3} +_{1} H^{1}$
$_{1} H^{2} +_{1} H^{3} \to_{2} H e^{4} +_{0} n^{1}$
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 :

10^{6} s

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10^{8} s

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10^{12} s

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10^{16} s

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Solution

The correct option is C $10^{12} s$
The mass defect $Δ m = 1.11 \times 10^{- 53}$ kg
The energy produced $E = Δ m C^{2}$
$E = 1.11 \times 10^{- 53} \times (3 \times 10^{8})^{2} = 10^{- 36} J$
For $10^{40}$ deuterons, the energy produced
$= 10^{40} \times 10^{- 36} J = 10^{4} J$
The average power radiated by star is $10^{16} W = 10^{16} J s$
The deuteron supply of the star is exhausted in a time of the order of $\frac{10^{16} J s}{10^{4} J} = 10^{12} s$

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Similar questions

Q. A star initially has

10^{40}

deuterons. It produces energy via the processes

_{1} H^{2} +_{1} H^{2} \to_{1} H^{3} + p

and

_{1} H^{2} +_{1} H^{3} \to_{2} H e^{4} + n

. If the average power radiated by the star is

10^{16}

W, the deuteron supply of the star is exhausted in a time of the order of

[The mass of nuclei are as follows:

M (H^{2}) = 2.014 a m u, M (n) = 1.008 a m u,

M (p) = 1.007 a m u, M (H e^{4}) = 4.001 a m u

}

Q. A star initially has

10^{40}

deuterons. It produces energy via the processes

_{1} H^{2} +_{1} H^{2} \to_{1} H^{3} + p and_{1} H^{2} +_{1} H^{3} \to_{2} {He}^{4} + n

. If the average power radiated by the star is

10^{16} W

, the deuteron supply of the star is exhausted in a time of the order of

11.6 \times 10^{x} s

. Then the value of

x

is -

Take m (_{1} H^{2}) = 2.0141 u, m (_{2} {He}^{4}) = 4.0026 u, m (_{1} p^{1}) = 1.0078 u, m (_{0} n^{1}) = 1.0086 u 1 u c^{2} = 931.5 MeV

Q. A star initially has

10^{40}

deuterons. It produces energy via the processes,

_{1} H^{2} +_{1} H^{2} \to_{1} H^{3} + p

_{1} H^{2} +_{1} H^{3} \to_{2} {He}^{4} + n

The masses of the nuclei are as follows,

m (_{1} H^{2}) = 2.014 amu; m (_{1} H^{3}) = 3.016 amu m (p) = 1.007 amu; m (n) = 1.008 amu m (_{2} {He}^{4}) = 4.001 amu

If the average power radiated by the star is

10^{16} W

, the deuteron supply of the star is exhausted in a time period of the order of
[Take

1 amu = 931.5 \frac{MeV}{c^{2}}

]

Q. A star initially has 10

^{40}

deuterons. It produces energy via the process

H^{2} +_{1} H^{2} \to_{1} H^{3} + p

and

_{1} H^{2} \to_{1} H^{3} \to_{2} H e^{4} + n

. If the average power radiated by the star is 10

^{16}

W, the deuteron supply of the star is exhausted in a time of the order of :

Q. A star initially has

10^{40}

deuterons. It produces energy via the processes

_{1}^{2} H +_{1}^{2} H \to_{1}^{3} H +_{0}^{1} p a n d_{1}^{2} H +_{1}^{3} H \to_{2}^{4} He + n

. If the average power radiated by the star is

10^{16}

W, the deuteron supply of the star is exhausted in a time of the order of

[Given M

(^{2} H) = 2.014 u, M (n) = 1.008 u, (p) = 1.008 u,

and

M (^{4} H e) = 4.001 u]

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A star has 1040 deuterons. It produces energy via the process1H2+1H2→1H3+1H11H2+1H3→2He4+0n1If the average power radiated by the star is 1016 W, then the deuteron supply of the star is exhausted in a time of the order of :