SATHEE: Nuclei - Result Question 70

Nuclei - Result Question 70

71. The activity of a radioactive sample is measured as $N_{0}$ counts per minute at $t = 0$ and $N_{0} / e$ counts per minute at $t = 5$ minutes. The time (in minutes) at which the activity reduces to half its value is

(a) $\log_{e} 2 / 5$

(b) $\frac{5}{\log_{e} 2}$

(d) $5 \log_{e} 2$

[2010]

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Answer:

Correct Answer: 71. (d)

Solution:

(d) $N = N_{0} e^{- λ t}$

Here, $t = 5$ minutes

$\frac{N_{0}}{e} = N_{0} \cdot e^{- 5 λ}$

$\Rightarrow 5 λ = 1$ ,

$λ = \frac{1}{5}$ ,

Now, $T_{1 / 2} = \frac{\ln 2}{λ} = 5 \ln 2$

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Nuclei - Result Question 70

71. The activity of a radioactive sample is measured as N0 counts per minute at t=0 and N0/e counts per minute at t=5 minutes. The time (in minutes) at which the activity reduces to half its value is

Answer:

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71. The activity of a radioactive sample is measured as $N_{0}$ counts per minute at $t = 0$ and $N_{0} / e$ counts per minute at $t = 5$ minutes. The time (in minutes) at which the activity reduces to half its value is