At a given instant there are 25% undecayed radioactive nuclei.After 10 sec the number of undecayed nuclei reduces to 6.25% the mean life of the nuclei is

14.43 s e c

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7.2 s e c

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1 s e c

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5 s e c

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Solution

The correct option is B $7.2 s e c$
In $10 s e c;$ number of nuclei has been reduced to one fourth $(25 % t o 6.25 %)$
Therefore it's half life is $T = 5 s e c$
$∴$ Mean life $T = \frac{T_{1 / 2}}{0.693} = \frac{5}{0.693} = 7.21 s e c$

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Similar questions

At a given instant there are 25% undecayed radioactive nuclei in a samplee. After 10 sec the number of undecayed nuclei reduces to 6.25%, the mean life of the nuclei is

Q. At a given instant these are

25 %

undecayed radioactive nuclei in a sample. After

10

second, the number of undecayed nuclei reduces to

12.5 %

calculate:
(a) Mean life of nuclei.
(b) The time in which the number of undecayed nuclei will further reduce to

6.25 %

of the reduced number.

Q. At a given instant there are 25 % undecayed radioactive nuclei in a sample. After 10 sec the number of undecayed nuclei remains 12.5 %. Calculate :
( i ) mean

-

life of the nuclei and
( ii ) The time in which the number of undecayed nuclears will further reduce to 6.25 % of the reduced number.

Q. At a given instant, there are 25% undecayed radioactive nuclei in a sample. After 10 s, the number of undecayed nuclei reduces to 12.5%. Calculate (a) mean life of the nuclei and (b) the time in which the number of decayed nuclei will further reduce to 6.25% of the reduced number

Q. At a given instant there are 25% undecayed radio-active nuclei in a sample. After 10 seconds the number of undecayed nuclei reduces to 12.5%. Calculate (i) mean-life of the nuclei, and (ii) the time in which the number of undecayed nuclei will further reduce to 6.25% of the reduced number.

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At a given instant there are 25% undecayed radioactive nuclei.After 10 sec the number of undecayed nuclei reduces to 6.25% the mean life of the nuclei is

The correct option is B 7.2secIn 10 sec; number of nuclei has been reduced to one fourth (25% to 6.25%)Therefore it's half life is T=5 sec∴ Mean life T=T1/20.693=50.693=7.21 sec

The correct option is B $7.2 s e c$
In $10 s e c;$ number of nuclei has been reduced to one fourth $(25 % t o 6.25 %)$
Therefore it's half life is $T = 5 s e c$
$∴$ Mean life $T = \frac{T_{1 / 2}}{0.693} = \frac{5}{0.693} = 7.21 s e c$