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Wave Optics

Question
CBSEENPH12039981
Wired Faculty App

The activity of a radioactive sample is measured as No counts per minute at t = 0 and No/e counts per minute at t = 5 min. The time (in minute) at which the activity reduces to half its value is

  • loge 2 /5

  • 5/ loge 2

  • 5 log10 2

  • 5 loge 2

Solution

D.

5 loge 2

Fraction remains after n half lives
straight N over straight N subscript straight o space equals space open parentheses 1 half close parentheses to the power of straight n space equals space open parentheses 1 half close parentheses to the power of straight t divided by straight T end exponent Given space space straight N space equals space straight N subscript straight o over straight e space rightwards double arrow space straight N subscript straight o over eN subscript straight o space equals space open parentheses 1 half close parentheses to the power of 5 divided by straight T end exponent Or space 1 over straight e space equals space open parentheses 1 half close parentheses to the power of 5 divided by straight T end exponent Taking space log space on space both space side comma space we space get log space 1 space minus space log space straight e space equals space 5 over straight T space log space 1 half minus 1 space equals space 5 over straight T space left parenthesis negative log space 2 thin space right parenthesis straight T space equals space log subscript straight e space 2 Now comma space let space straight t apostrophe space be space the space time space after space which space activity space reduces space to space half open parentheses 1 half close parentheses space equals open parentheses space 1 half close parentheses to the power of straight t apostrophe divided by 5 space log subscript straight e 2 end exponent straight t apostrophe space equals space 5 space log subscript straight e space 2

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