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Physics-

General

Easy

Question

An atomic power nuclear reactor can deliver 300 MW. The energy released due to fission of each nucleus of uranium atom $U to the power of 238 end exponent$ is 170 MeV. The number of uranium atoms fissioned per hour will be

$30 \times 1 0^{25}$
$4 \times 1 0^{22}$
$10 \times 1 0^{20}$
$5 \times 1 0^{15}$

The correct answer is: $4 cross times 1 0 to the power of 22 end exponent$

$P equals fraction numerator n E over denominator t end fraction rightwards double arrow 300 cross times 1 0 to the power of 6 end exponent equals fraction numerator n cross times 170 cross times 1 0 to the power of 6 end exponent cross times 1.6 cross times 1 0 to the power of negative 19 end exponent over denominator t end fraction$
\ Number of atoms per sec $blank fraction numerator n over denominator t end fraction equals 1.102 cross times 1 0 to the power of 19 end exponent$
Number of atoms per hour = 1.02 $cross times$ 10¹⁹ $cross times$ 3600
= 3.97 $cross times$ 10²².

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