Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

At some instant, the number of radioactive atoms in a sample is $N_0$ and after time $t$ , the number decreases to $N$ . It is found that the graphical representation $\ln N$ versus $t$ along the $y$ and $x$ axis respectively is a straight line. Then the slope of this line is:

$\lambda$

$-\lambda$

$\lambda^{-1}$

$-\lambda^{-1}$

Step-by-Step Solution

Identify the Decay Law: Radioactive decay follows first-order kinetics. The number of undecayed nuclei $N$ at time $t$ is given by the equation $N = N_0 e^{-\lambda t}$ , where $\lambda$ is the decay constant .
Linearize the Equation: To interpret the graph, take the natural logarithm ( $\ln$ ) of both sides: $\ln N = \ln(N_0 e^{-\lambda t})$ $\ln N = \ln N_0 + \ln(e^{-\lambda t})$ $\ln N = -\lambda t + \ln N_0$
Compare with Straight Line Equation: This equation is in the form $y = mx + c$ , which represents a straight line. Matching the terms:

$y = \ln N$
$x = t$
$m$ (slope) = $-\lambda$
$c$ (intercept) = $\ln N_0$

Conclusion: The slope of the graph of $\ln N$ versus $t$ is $-\lambda$ .

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