If N_t = N_o e^{-lambda t},then the number of | vedclass

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

Question

(C) The number of undecayed nuclei at any time $t$ is given by $N_t = N_o e^{-\lambda t}$.At time $t_1$,the number of undecayed nuclei is $N(t_1) = N_

Vedclass · Accepted Answer

$N_o [e^{-\lambda t_1} - e^{-\lambda t_2}]$

Vedclass · Answer

(C) The number of undecayed nuclei at any time $t$ is given by $N_t = N_o e^{-\lambda t}$.At time $t_1$,the number of undecayed nuclei is $N(t_1) = N_o e^{-\lambda t_1}$.At time $t_2$,the number of undecayed nuclei is $N(t_2) = N_o e^{-\lambda t_2}$.The number of nuclei that have decayed between time $t_1$ and $t_2$ is the difference between the number of undecayed nuclei at $t_1$ and $t_2$:$\Delta N = N(t_1) - N(t_2)$Substituting the expressions:$\Delta N = N_o e^{-\lambda t_1} - N_o e^{-\lambda t_2}$Taking $N_o$ as a common factor:$\Delta N = N_o [e^{-\lambda t_1} - e^{-\lambda t_2}]$

If $N_t = N_o e^{-\lambda t}$,then the number of nuclei decaying between time $t_1$ and $t_2$ $(t_2 > t_1)$ is given by:

Similar Questions

After two hours, one-sixteenth of the starting amount of a certain radioactive isotope remained undecayed. The half-life of the isotope is:

State the relation between average life and decay constant.

In a radioactive disintegration,the ratio of the initial number of atoms to the number of atoms present at an instant of time equal to its mean life is:

If the half-life of a radioactive nucleus is $3$ days, nearly what fraction of the initial number of nuclei will decay on the third day? (Given, $\sqrt[3]{0.25} \approx 0.63$)

Calculate the time (in $minutes$) interval between $33\%$ decay and $67\%$ decay if the half-life of a substance is $20\, minutes$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module