In saloons, there is always a characteristics smell due to ammonia-based chemicals used in hair dyes

English

Gujarati

Hindi

13.Nuclei

hard

In saloons, there is always a characteristics smell due to the ammonia-based chemicals used in hair dyes and other products. Assume the initial concentration of ammonia molecules to be $1000 \,molecules/ m ^3$. Due to air ventilation, the number of molecules leaving in one minute is one tenth of the molecules present at the start of that minute. How long will it take for the concentration of ammonia molecules to reach $1 \,molecule / m ^3$ ?

A

$7$ minutes

B

$70$ minutes

C

$100$ minutes

D

Very long time which cannot be calculated.

(KVPY-2021)

Solution

(B)

The given situation is like an exponential decay.

$\therefore N=N_0 e^{-\lambda t}$

Here $\frac{9}{10} 1000=1000 e ^{-\lambda s}$

$\Rightarrow \ln (0.9)=-\lambda$

$\Rightarrow \lambda=0.1$

Now $1=1000 e ^{-0.1 t}$

$\Rightarrow \ln (0.001)=-0.1 t$

$\Rightarrow \frac{6.9}{0.1}= t \Rightarrow t \approx 70 \,\min$

Standard 12

Physics

Share

Similar Questions

A radioactive material decays by simultaneous emission of two particles with respective half lives $1620$ and $810$ years. The time (in years) after which one- fourth of the material remains is

medium

(AIIMS-2008)

For a certain radioactive process the graph between $In\, {R}$ and ${t}\,({sec})$ is obtained as shown in the figure. Then the value of half life for the unknown radioactive material is approximately $….\,{sec}.$

medium

(JEE MAIN-2021)

A radioactive isotope has a half-life of $T$ years. How long will it take the activity to reduce to $(a)$ $3.125\% $ $(b)$ $1\% $ of its original value?

medium

In a radioactive decay chain, ${ }_{90}^{232} Th$ nucleus decays to ${ }_{82}^{212} Pb$ nucleus. Let $N _\alpha$ and $N _\beta$ be the number of $\alpha$ and $\beta^{-}$particles, respectively, emitted in this decay process. Which of the following statements is (are) true?

$(A)$ $N _\alpha=5$ $(B)$ $N _\alpha=6$ $(C)$ $N _\beta=2$ $(D)$ $N _\beta=4$

medium

(IIT-2018)

In a radioactive decay process, the activity is defined as $A=-\frac{\mathrm{d} N}{\mathrm{~d} t}$, where $N(t)$ is the number of radioactive nuclei at time $t$. Two radioactive sources, $S_1$ and $S_2$ have same activity at time $t=0$. At a later time, the activities of $S_1$ and $S_2$ are $A_1$ and $A_2$, respectively. When $S_1$ and $S_2$ have just completed their $3^{\text {rd }}$ and $7^{\text {th }}$ half-lives, respectively, the ratio $A_1 / A_2$ is. . . . . . .

medium

(IIT-2023)