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13.Nuclei
hard
રેડિયો એક્ટિવ તત્વનો અર્ધઆયુ $20$ મિનિટ નો છે. $33\%$ અને $67\%$ વિભંજન વચ્ચેના સમય ......... મિનિટ
A
$10$
B
$20$
C
$30$
D
$40$
(AIIMS-2000)
Solution
(b) $\lambda = \frac{{0.693}}{{{T_{1/2}}}} = \frac{{0.693}}{{20}} = 0.03465$
Now time of decay $t = \frac{{2.303}}{\lambda }\log \frac{{{N_0}}}{N}$
$ \Rightarrow {t_1} = \frac{{2.303}}{{0.03465}}\log \frac{{100}}{{67}} = 11.6\, min $
and ${t_2} = \frac{{2.303}}{{0.03465}}\log \frac{{100}}{{33}} = 32\,min$
Thus time difference between points of time
$= t_1 -t_2 =32 -11.6 = 20.4\, min \, \approx \, 20\, min.$
Standard 12
Physics
