MHT CET · Physics · Nuclear Physics
A radioactive element has a rate of disintegration 10,000 disintegrations per minute at a particular instant. After four minutes it becomes 2500 disintegrations per minute. The decay constant per minute is
- A
- B
- C
- D
Answer & Solution
Correct Answer
(B)
Step-by-step Solution
Detailed explanation
\(\frac{N}{N_0}=e^{-\lambda t}\)
\(\therefore\) \(\frac{2500}{10000}=e^{-\lambda \times 4}\)
\(\therefore\) \(\frac{1}{4}=e^{-4 \lambda} \)
\(\therefore\) \( e^{4 \lambda}=4 \)
\(\therefore\) \( 4 \lambda=\log _e 4 \)
\(\therefore\) \( 4 \lambda=\log _e 2^2 \)
\(\therefore\) \( 4 \lambda=2 \log _e 2 \)
\(\therefore\) \( \lambda=\frac{2}{4} \log _e 2 \)
\(\therefore\) \( \lambda=0.5 \log _e 2\)
\(\therefore\) \(\frac{2500}{10000}=e^{-\lambda \times 4}\)
\(\therefore\) \(\frac{1}{4}=e^{-4 \lambda} \)
\(\therefore\) \( e^{4 \lambda}=4 \)
\(\therefore\) \( 4 \lambda=\log _e 4 \)
\(\therefore\) \( 4 \lambda=\log _e 2^2 \)
\(\therefore\) \( 4 \lambda=2 \log _e 2 \)
\(\therefore\) \( \lambda=\frac{2}{4} \log _e 2 \)
\(\therefore\) \( \lambda=0.5 \log _e 2\)
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