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Physics Class 12th Nuclei Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is:

(Take log₁₀ 1.88 = 0.274)

Option 1 -

0.02 min^-1

Option 2 -

2.7 min^-1

Option 3 -

0.063 min^-1

Option 4 -

6.3 min^-1

3 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

3 months ago

Correct Option - 3

Detailed Solution:

$A = A o e^{- λ t}$

$\Rightarrow 2250 = 4250 e^{- λ t}$

$\Rightarrow λ \times 10 \times 0.434 = 0.274$

$\Rightarrow λ = \frac{0.274}{10 \times 0.434} = 0.63 m i n^{- 1}$

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The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is:

(Take log₁₀ 1.88 = 0.274)

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The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is: (Take log10 1.88 = 0.274)

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The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is:

(Take log₁₀ 1.88 = 0.274)