Question

In a sample of uranium the ratio of ${}^{238}\text{U}$ and ${}^{235}\text{U}$ by weight is 93 to 7%. The half-life of ${}^{238}\text{U}$ and ${}^{235}\text{U}$ are $4.5\times {10}^9$ and $7.13\times {10}^8$ year respectively. The contribution of radioactivity by both separately after a time interval of ${10}^8$ year if 100 g sample is taken is :

• A. $[4.56\times {10}^{13}dpyof{}^{238}\text{U},1.56\times {10}^{13}dpyof{}^{235}\text{U}]$
• B. $[3.56\times {10}^{13}dpyof{}^{238}\text{U},1.56\times {10}^{13}dpyof{}^{235}\text{U}]$
• C. $[5.56\times {10}^{13}dpyof{}^{238}\text{U},1.56\times {10}^{13}dpyof{}^{235}\text{U}]$
• D. $[6.56\times {10}^{13}dpyof{}^{238}\text{U},1.56\times {10}^{13}dpyof{}^{235}\text{U}]$

Answer 1

The correct option is B $[3.56\times {10}^{13}dpyof{}^{238}\text{U},1.56\times {10}^{13}dpyof{}^{235}\text{U}]$

100 g of sample contains 93 g of U-238 and 7 g of U-235.
For U-238, the number of nuclei present in 93 g of sample are
$\frac{93}{238}\times 6.023\times {10}^{23}=2.35\times {10}^{23}$
The decay constant is $\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{4.5\times {10}^9}=1.54\times {10}^{-10}yr$
The contribution to the radioactivity is $\lambda N=1.54\times {10}^{-10}\times 2.35\times {10}^{23}=3.56\times {10}^{13}dpy$
For U-235, the number of nuclei present in 7 g of sample are
$\frac{7}{235}\times 6.023\times {10}^{23}=1.79\times {10}^{22}$
The decay constant is $\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{7.13\times {10}^8}=9.72\times {10}^{-10}yr$
The contribution to the radioactivity is $\lambda N=9.72\times {10}^{-10}\times 1.79\times {10}^{22}=1.56\times {10}^{13}dpy$