A) \[333\]
B) \[100\]
C) \[250\]
D) \[125\]
Correct Answer: A
Solution :
Rate of dissociation \[\frac{d[N]}{dt}=\lambda {{N}_{1}}=1000\] \[{{N}_{1}}\] atom \[t/2\] = present in 1 hour \[{{N}_{2}}\] atom t = present in 3 hour \[\lambda =\frac{2.303}{t/2}\log \frac{{{N}_{0}}}{{{N}_{1}}}\] \[\lambda =\frac{2.303}{t}\log \frac{{{N}_{0}}}{{{N}_{2}}}\] \[\frac{2.303}{1}\log \frac{{{N}_{0}}}{{{N}_{1}}}=\frac{2.303}{3}\log \frac{{{N}_{0}}}{{{N}_{2}}}\] \[\frac{{{N}_{1}}}{{{N}_{2}}}=\frac{3}{1}\] or \[\frac{\lambda {{N}_{1}}}{\lambda {{N}_{2}}}=\frac{3}{1}\] \[\lambda {{N}_{1}}=3\lambda {{N}_{2}}\] or \[1000\,per\,\sec =3\lambda {{N}_{2}}\] \[\lambda {{N}_{2}}=\frac{1000}{3}=333\sec \]
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