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Physics Nuclei Physics Class 12th Nuclei

For a ceratin radioactive process the graph between $l n (R)$ and t(sec) is obtained as shown in the figure. Then the value of half life for the unknown radioactive material is approximately:

Option 1 -

2.62 sec

Option 2 -

9.15 sec

Option 3 -

4.62 sec

Option 4 -

6.93 sec

2 Views | Posted 3 months ago

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

3 months ago

Correct Option - 3

Detailed Solution:

$R = - \frac{d N}{d t} = λ N = λ N_{0} e^{- λ t} \Rightarrow l n (R) = l n (λ N_{0}) - λ t, s o$

$λ = t a n θ = \frac{6}{40} = \frac{3}{20} s^{- 1} \Rightarrow$ Slope of graph

$\Rightarrow t_{\frac{1}{2}} = \frac{0.693}{λ} = \frac{0.693 \times 20}{3} = 4.62 s$

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For a ceratin radioactive process the graph between $l n (R)$ and t(sec) is obtained as shown in the figure. Then the value of half life for the unknown radioactive material is approximately:

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For a ceratin radioactive process the graph between $l n (R)$ and t(sec) is obtained as shown in the figure. Then the value of half life for the unknown radioactive material is approximately: