Explanation:
Here, \({R_0} = 4.0 \times {10^6}\) disintegration \(/ {s}\),
\(R = 1.0 \times {10^6}{\text{ disintegration/}}s\)
\({t}=20\) hours, \({t}^{\prime}=100\) hours
As, \(\dfrac{{R}}{{R}_{0}}=\left(\dfrac{1}{2}\right)^{n}\)
\(\therefore \dfrac{1.0 \times 10^{6}}{4.0 \times 10^{6}}=\left(\dfrac{1}{2}\right)^{n}\)
\(\therefore\left(\dfrac{1}{2}\right)^{2}=\left(\dfrac{1}{2}\right)^{n}\)
\(\therefore {n}=2\)
But, \({T}_{1 / 2}=\dfrac{{t}}{{n}}=\dfrac{20}{2}=10\) hours
Now, \({n}^{\prime}=\dfrac{{t}^{\prime}}{{T}_{1 / 2}}=\dfrac{100}{10}=10\)
\(\therefore \quad {R}^{\prime}={R}_{0}\left(\dfrac{1}{2}\right)^{{n}^{\prime}}\)
\(=4.0 \times 10^{6} \times\left(\dfrac{1}{2}\right)^{10}\)
\(=\dfrac{10^{6}}{2^{8}}\)
\(=\dfrac{10}{2^{2}} \times \dfrac{10}{2^{2}} \times \dfrac{10}{2^{2}} \times \dfrac{10}{2^{2}} \times 10^{2}\)
\(=2.5 \times 2.5 \times 2.5 \times 2.5 \times 10^{2}\)
\(=39.0625 \times 10^{2}\)
\(=3.91 \times 10^{3}\) disintegration \(/ {s}\)
Comparing with ' \({N}\) ' \(\times 10^{3}\) disintegration \(/ {s}\),
\(N = 3.91\)