SATHEE: Atoms And Nuclei Question 407

Atoms And Nuclei Question 407

Question: The proton -proton mechanism that accounts for energy production in the sun releases $26.7 M e V$ energy for each event. In this process, protons fuse to form an alpha particle $(^{4} H e)$ . At what rate $\frac{d m}{d t}$ is hydrogen being consumed in the core of the sun by the p-p cycle. Power of sun is $3.90 \times 10^{26} W$ .

Options:

A) $1.6 \times 10^{10} k g / s$

B) $2.3 \times 10^{9} k g / s$

C) $6.2 \times 10^{11}, k g / s$

D) $5.5 \times 10^{10}, k g / s$

Show Answer

Answer:

Correct Answer: C

Solution:

The rate $d m / d t$ 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 $26.2, M e V = 4.20 \times 10^{- 12} J$

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 $\frac{d m}{d t} = \frac{Δ m}{Δ E} P = \frac{4 (1.67 \times 10^{- 27})}{4.20 \times 10^{- 12}} \times (3.90 \times 10^{26})$ $= 6.2 \times 10^{11} k g / s$

Atoms And Nuclei Question 407

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Atoms And Nuclei Question 407

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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