Question

The half life of 32p is 14.2 days. how long would it take a solution containing 42000 dpm to decay to 500 dpm?​

Answer 1

Answer:

The specific activity of a radionuclide is its activity per kg of the radioactive material.

$1kg=1000g </p><p>31</p><p> P= </p><p>31</p><p>1000</p><p> </p><p> N </p><p>0</p><p> </p><p> atoms of </p><p>31</p><p> P$

$But </p><p>32</p><p> P content is 1ppm(1 atm </p><p>32</p><p> P out of 10 </p><p>6</p><p> atoms)$

$∴ </p><p>32</p><p> P= </p><p>31</p><p>1000N </p><p>0</p><p> </p><p> </p><p> </p><p> × </p><p>10 </p><p>6</p><p> </p><p>1</p><p> </p><p> = </p><p>31×10 </p><p>6</p><p> </p><p>1000×6.02×10 </p><p>23</p><p> </p><p> </p><p> </p><p>=1.942×10 </p><p>19</p><p> atoms of </p><p>32</p><p> P</p><p>also − </p><p>dt</p><p>dN</p><p> </p><p> =λN, (λ= </p><p>T </p><p>50</p><p> </p><p> </p><p>0.693</p><p> </p><p> )</p><p>− </p><p>dt</p><p>dN</p><p> </p><p> = </p><p>14.3×24×3600</p><p>0.693</p><p> </p><p> ×1.942×10 </p><p>19</p><p> dps</p><p>=1.089×10 </p><p>13</p><p> s </p><p>−1</p><p>$

hence, specific activity(number of curies in 1 kg of 32P)

$3.7×10 </p><p>10</p><p> </p><p>1.089×10 </p><p>13</p><p> </p><p> </p><p> =294.32Cikg </p><p>−1$