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  • The power obtained in a reactor using disintegration is <img alt="1000 \mathrm{~kW}" src="/latex-image/?1000%20%5Cmat

The power obtained in a reactor using \mathrm{U}^{235} disintegration is 1000 \mathrm{~kW} .The mass decay of  \mathrm{U}^{235} per hour is: 

Option: 1

10 microgram


Option: 2

20 microgram


Option: 3

40 microgram


Option: 4

1 microgram


Answers (1)

best_answer

According to Einstein’s mass energy relation

\mathrm{E}=\mathrm{mc}^2 \quad \text { or } \quad \mathrm{m}=\frac{\mathrm{E}}{\mathrm{c}^2}

\text { Mass decay per second }=\frac{\Delta \mathrm{m}}{\Delta \mathrm{t}}=\frac{1}{\mathrm{c}^2} \frac{\Delta \mathrm{E}}{\Delta \mathrm{t}}=\frac{\mathrm{P}}{\mathrm{c}^2}=\frac{1000 \times 10^3 \mathrm{~W}}{\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)^2}=\frac{10^6}{9 \times 10^{16}} \mathrm{~kg} / \mathrm{s}

\text { Mass decay per hour }=\frac{\Delta \mathrm{m}}{\Delta \mathrm{t}} \times 60 \times 60=\left(\frac{10^6}{9 \times 10^{16}} \mathrm{~kg} / \mathrm{s}\right)(3600 \mathrm{~s})=4 \times 10^{-8} \mathrm{~kg}=40 \times 10^{-6} \mathrm{~g}=40 \mu \mathrm{g}

Posted by

Divya Prakash Singh

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