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Wave Optics

Question
CBSEENPH12040080
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Two radioactive substance A and B have decay constants 5λ and λ respectively. At t = 0 they have the same number of nuclei. The ratio of a number of nuclei of A t\o those of B will be open parentheses 1 over straight e close parentheses squared after a time interval:

  • 1/ 4λ

  • 4λ

  • 2λ

  • 1/2λ

Solution

D.

1/2λ

Number of nuclei remained after time t can be written as
N = Noe-λt
where No is initial number of nuclei of both the substances.
straight N subscript 1 space equals space straight N subscript straight o straight e to the power of negative 5 λt space end exponent space space... space left parenthesis straight i right parenthesis straight N subscript 2 space equals space straight N subscript straight o straight e to the power of negative λt end exponent space space space space space.... space left parenthesis ii right parenthesis Dividing space eq. space left parenthesis straight i right parenthesis space by space Eq space left parenthesis ii right parenthesis space obtain straight N subscript 1 over straight N subscript 2 space equals space open parentheses 1 over straight e close parentheses squared space equals space open parentheses 1 over straight e squared close parentheses hence comma space 1 over straight e squared space equals space 1 over straight e to the power of 4 λt end exponent Comparing space the space powers comma space we space get 2 space equals space 4 λt straight t space equals space fraction numerator 2 over denominator 4 straight lambda end fraction space equals space fraction numerator 1 over denominator 2 straight lambda end fractionstraight N subscript 1 space equals space straight N subscript straight o straight e to the power of negative 5 λt space end exponent space space... space left parenthesis straight i right parenthesis straight N subscript 2 space equals space straight N subscript straight o straight e to the power of negative λt end exponent space space space space space.... space left parenthesis ii right parenthesis Dividing space eq. space left parenthesis straight i right parenthesis space by space Eq space left parenthesis ii right parenthesis space obtain straight N subscript 1 over straight N subscript 2 space equals space open parentheses 1 over straight e close parentheses squared space equals space open parentheses 1 over straight e squared close parentheses hence comma space 1 over straight e squared space equals space 1 over straight e to the power of 4 λt end exponent Comparing space the space powers comma space we space get 2 space equals space 4 λt straight t space equals space fraction numerator 2 over denominator 4 straight lambda end fraction space equals space fraction numerator 1 over denominator 2 straight lambda end fraction

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