A fresh radioactive sample is given at t = 0. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $N_0$ be the initial number of radioactive nuclei.At time $t_1$,the decay fraction is $\frac{1}{5}$,so the remaining fraction is $1 - \frac{1}

Vedclass · Accepted Answer

$\frac{t_2 - t_1}{\ln 4}$

Vedclass · Answer

(B) Let $N_0$ be the initial number of radioactive nuclei.At time $t_1$,the decay fraction is $\frac{1}{5}$,so the remaining fraction is $1 - \frac{1}{5} = \frac{4}{5}$. Thus,$N(t_1) = N_0 \frac{4}{5}$.At time $t_2$,the decay fraction is $\frac{4}{5}$,so the remaining fraction is $1 - \frac{4}{5} = \frac{1}{5}$. Thus,$N(t_2) = N_0 \frac{1}{5}$.The decay law is $N(t) = N_0 e^{-\lambda t}$.For $t_1$: $\frac{4}{5} = e^{-\lambda t_1} \implies \ln(\frac{4}{5}) = -\lambda t_1$.For $t_2$: $\frac{1}{5} = e^{-\lambda t_2} \implies \ln(\frac{1}{5}) = -\lambda t_2$.Subtracting the two equations: $\ln(\frac{1}{5}) - \ln(\frac{4}{5}) = -\lambda t_2 - (-\lambda t_1) = -\lambda(t_2 - t_1)$.$\ln(\frac{1/5}{4/5}) = -\lambda(t_2 - t_1) \implies \ln(\frac{1}{4}) = -\lambda(t_2 - t_1)$.$-\ln(4) = -\lambda(t_2 - t_1) \implies \lambda = \frac{\ln 4}{t_2 - t_1}$.Mean life $T_m = \frac{1}{\lambda} = \frac{t_2 - t_1}{\ln 4}$.

$A$ fresh radioactive sample is given at $t = 0$. Its decay fraction is $\frac{1}{5}$ at $t_1$ instant and $\frac{4}{5}$ at $t_2$ instant. Its mean life is

Similar Questions

Considering only the principal values of inverse functions,the set $A = \{x \geq 0, \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}\}$

Four resistors $A, B, C$ and $D$ form a Wheatstone bridge. The bridge is balanced when $C = 100 \ \Omega$. If $A$ and $B$ are interchanged,the bridge balances for $C = 121 \ \Omega$. The value of $D$ is (in $\Omega$)

The mean of a set of observations is $\bar{x}$. If each observation is divided by $\alpha$ $(\alpha \neq 0)$ and then increased by $10$,what is the mean of the new set?

In a compound microscope,the focal lengths of two lenses are $1.5 \, cm$ and $6.25 \, cm$. An object is placed at $2 \, cm$ from the objective lens and the final image is formed at $25 \, cm$ from the eye lens. The distance between the two lenses is.......$cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module