A radioactive β –emission. A detector records n β –particles in 2s and by next 2s (accumulatively) it records 1.1 n β –particle. Number of β –particles recorded by detector after a long time, is
A radioactive β –emission. A detector records n β –particles in 2s and by next 2s (accumulatively) it records 1.1 n β –particle. Number of β –particles recorded by detector after a long time, is
Option 1 - <p>11n/10<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 2 - <p>12n/10<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 3 - <p>2n</p>
Option 4 - <p>10n/9</p>
1 Views|Posted 5 months ago
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Answered by
5 months ago
Correct Option - 4
Detailed Solution:
N = [n + n/10 + n/100 + .]
= n/ (1 - 1/10) = 10/9
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N = N0e–λt
->t = λN = N0/e
So number of nuclei decayed = N0 – N
= N0 (1- 1/e)
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Physics Nuclei 2025
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