KCET · Physics · Nuclear Physics
The half life of tritium is \( 12.5 \) years. What mass of tritium of initial mass \( 64 \mathrm{~g} \) will remain
undecayed after \( 50 \) years?
- A \( 32 \mathrm{mg} \)
- B \( 8 \mathrm{mg} \)
- C \( 16 \mathrm{mg} \)
- D \( 4 \mathrm{mg} \)
Answer & Solution
Correct Answer
(D) \( 4 \mathrm{mg} \)
Step-by-step Solution
Detailed explanation
Given, half-life of tritium \(=12.5\) years; initial mass \(=64 \mathrm{mg} ;\) time, \(\mathrm{t}=50\) years
We know that
\(N=N_{0} e^{-\lambda t}\)
and \(\lambda=\frac{0.693}{T_{1 / 2}}=\frac{0.693}{12.5 \text { years }}\)
\(\therefore \frac{N}{N_{0}}=e^{\frac{-0.693}{12.5} \times 50}\)
\(\Rightarrow \frac{N}{N_{0}}=e^{-2.772}=0.0625\)
\(\Rightarrow N=N_{0} \times 0.0625=64 \times 0.0625 \mathrm{mg}\)
\(\Rightarrow N=4 \mathrm{mg}\)
Thus, 4 mg of tritium will remain undecayed after 50 years
We know that
\(N=N_{0} e^{-\lambda t}\)
and \(\lambda=\frac{0.693}{T_{1 / 2}}=\frac{0.693}{12.5 \text { years }}\)
\(\therefore \frac{N}{N_{0}}=e^{\frac{-0.693}{12.5} \times 50}\)
\(\Rightarrow \frac{N}{N_{0}}=e^{-2.772}=0.0625\)
\(\Rightarrow N=N_{0} \times 0.0625=64 \times 0.0625 \mathrm{mg}\)
\(\Rightarrow N=4 \mathrm{mg}\)
Thus, 4 mg of tritium will remain undecayed after 50 years
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