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Wave Optics

Question
CBSEENPH12039913
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The power obtained in a reactor using U235 disintegration is  1000kW. The mass decay of U235 per hour is

  • 20 μg

  • 40 μg

  • 1 μg

  • 10 μg

Solution

B.

40 μg

Let assume power P = 1000 W
Energy per hour = 1000 x 3600 J
Energy per fission = 200 MeV
= 200 x 1.6 x 10-13 J
therefore number per hour 
straight n space equals space fraction numerator 1000 space straight x space 3600 over denominator 200 space straight x space 1.6 space straight x space 10 to the power of negative 13 end exponent 0 end fraction Number space of space mole space per space hour space space equals space straight n over straight N therefore comma space Mass space per space hour space equals space straight n over straight N space straight x space 235 equals space fraction numerator 1000 space straight x space 3600 space straight x space 235 over denominator 200 space straight x space 1.6 space straight x space 10 to the power of negative 13 end exponent space straight x space 6.02 space straight x space 10 to the power of 23 end fraction equals space 43.9 space straight x space 10 to the power of negative 6 space end exponent straight g
The nearest value is 40 μg, Hence option b is correct.

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