Question

A graph is plotted between $\log N$ vs. time for the first order reaction. Slope and intercept of the graph are-
A. $ - \dfrac{\lambda }{{2.303}}, \log {N_0}$
B. $\dfrac{\lambda }{{2.303}}, \log {N_0}$
C. $ - \dfrac{\lambda }{{2.303}}, {N_0}$
D. $\dfrac{\lambda }{{2.303}}, {N_0}$

Answer 1

Hint: Use the radioactive decay law which is given as-
$ \Rightarrow N = {N_0}{e^{ - \lambda t}}$ Where N is the number of nuclei in a sample, ${N_0}$ is the original number of nuclei in the sample at a time t and $\lambda $ is the constant of proportionality or radioactive decay constant, e is exponential. Then solve this equation to get the intercept and slope by taking log both side and comparing the obtained equation to straight line equation-
$ \Rightarrow y = mx + c$ where m is the slope and x is variable and c is the intercept.

Step-by-Step Explanation-
From radioactive decay law we know that the radioactive decay per unit time is directly proportional to the number of nuclei or radioactive compounds. It is given as-
$ \Rightarrow N = {N_0}{e^{ - \lambda t}}$ -- (i)
Where N is the number of nuclei in a sample, ${N_0}$ is the original number of nuclei in the sample at a time t and $\lambda $ is the constant of proportionality or radioactive decay constant, e is exponential.
On taking log both sides in eq. (i) we get,
$ \Rightarrow \log N = \log \left( {{N_0}{e^{ - \lambda t}}} \right)$
We know that $\log mn = \log m + \log n$
So applying this rule we get,
$ \Rightarrow \log N = \log {e^{ - \lambda t}} + \log {N_0}$
Now we also know that,
$\log {m^n} = n\log m$
On applying this rule we get,
$ \Rightarrow \log N = - \lambda t\log e + \log {N_0}$
Here the base of log is $10$ and we know that,
$ \Rightarrow {\log _{10}}e = \dfrac{1}{{{{\log }_e}10}} = \dfrac{1}{{2.303}}$
So putting this value in the equation we get,
$ \Rightarrow \log N = \dfrac{{ - \lambda t}}{{2.303}} + \log {N_0}$ -- (ii)
Now we know the straight line equation is
$ \Rightarrow y = mx + c$ where m is the slope and x is variable and c is the intercept.
On comparing the straight line equation with eq. (ii) we het,
Slope= $\dfrac{{ - \lambda t}}{{2303}}$
And Intercept=$\log {N_0}$
So slope and intercept of the graph is plotted between $\log N$ vs. time for the first order reaction are $\dfrac{{ - \lambda t}}{{2303}}$and $\log {N_0}$.

Answer- Hence the correct answer is A.

Note: Radioactive decay is the process in which the nucleus of an unstable atom loses energy by emitting radiation to stable the isotopes.
- This isotope transforms into another element until it obtains a stable nucleus due to which a series of elements are formed. This series is called the decay series.
- Although some elements always remain radioactive and are not found in a stable form in nature like uranium.