A radioactive simple is undergoing decay. At any time t1, its activity is A and another time t2, the activity is A5. What is the average life time for the sample?

A=A0e−λt1 [Radio active decay law] A5=A0e−λ (t2−t1) ⇒ln5=λ (t2−t1) ⇒Average life=1λ= (t2−t1)ln5

A radioactive simple is undergoing decay. At any time t₁, its activity is A and another time t₂, the activity is $\frac{A}{5} .$ What is the average life time for the sample?

Option 1 - <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>−</mo> <msub> <mrow> <mi>t</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <mi>I</mi> <mi>n</mi> <mn>5</mn> </mrow> </mfrac> </mrow> </math>

Option 2 - <math> <mrow> <mfrac> <mrow> <mi>I</mi> <mi>n</mi> <mrow> <mo>(</mo> <mrow> <msub> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>t</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math>

Option 3 - <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>t</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>−</mo> <msub> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <mi>I</mi> <mi>n</mi> <mn>5</mn> </mrow> </mfrac> </mrow> </math>

Option 4 - <math> <mrow> <mfrac> <mrow> <mi>I</mi> <mi>n</mi> <mn>5</mn> </mrow> <mrow> <msub> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>−</mo> <msub> <mrow> <mi>t</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> </mrow> </math>

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Answered by

Payal Gupta | Contributor-Level 10

8 months ago

Correct Option - 1

Detailed Solution:

$A = A_{0} e^{- λ t_{1}}$ [Radio active decay law]

$\frac{A}{5} = A_{0} e^{- λ (t_{2} - t_{1})}$