In the given graph aside, What does the slope of the curve gives? Find the value of A. Find the Value of B.

In the given graph aside,

1. What does the slope of the curve give?
2. Find the value of A.
3. Find the Value of B.

Solution:

(a) The slope of curve gives us the value of disintegration constant or decay constant.

(b) From the curve we know that:

t = A

and $\ln \left( {\frac{{{N_0}}}{N}} \right) = 0.693$

We Know: N = N₀e^-λt

Or, $\frac{N}{{{N_0}}} = {e^{ - \lambda t}}$

$or,\frac{{{N_0}}}{N} = \frac{1}{{{e^{ - \lambda t}}}}$

$or,\frac{{{N_0}}}{N} = {e^{\lambda t}}$

$or,\ln \left( {\frac{{{N_0}}}{N}} \right) = \lambda t$……(i)

Here, t=A, then

$0.693 = \lambda A$

$\therefore A = \frac{{0.693}}{\lambda }$

i.e., Value of A represents half-life (T_h) of radioactive sample in graph.

(c) From the curve:

$\ln \left( {\frac{{{N_0}}}{N}} \right) = B$

And t = 2T_h

We Know, From Equation (i)

$\ln \left( {\frac{{{N_0}}}{N}} \right) = \lambda t$

B = λt=λ.2T_h=2λT_h

If T_h represent half-life, then,

B = 2λ.$\frac{{0.693}}{\lambda }$ = 2 × 0.693 = 1.386