Let r be the center to center distance between the alpha particles and the nucleus (Z = 89). When the alpha particle is at the stopping point, then
\(K=\frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { (Ze)(2e) }{ r } \)
(or) \(K=\frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { 2{ Ze }^{ 2 } }{ K } \)
\(=\frac { 9\times { 10 }^{ 9 }\times 2\times 80{ e }^{ 2 } }{ 4.5MeV } \)
\(=\frac { 9\times { 10 }^{ 9 }\times 2\times 80\times (1.6\times { 10 }^{ -19 })^{ 2 } }{ 4.5\times { 10 }^{ 6 }\times 1.6\times { 10 }^{ -19 }J } \)
\(=\frac { 9\times 160\times 1.6 }{ 4.5 } \times { 10 }^{ -16 }=512\times { 10 }^{ -16 }m\)
= 5.12 x 10^-14 m

\(\frac { R }{ { R }_{ 0 } } ={ \left( \frac { 1 }{ 2 } \right) }^{ n }\)
\(\frac { 3.125 }{ 100 } ={ \left( \frac { 1 }{ 2 } \right) }^{ n }\)
\(\frac { 1 }{ 32 } ={ \left( \frac { 1 }{ 2 } \right) }^{ n }={ \left( \frac { 1 }{ 2 } \right) }^{ 5 }\)
\(n=5,\frac { t }{ T } =5\)
t = 5T

Average life, T_α = \(\frac { 1 }{ \lambda } \)
Now, N = \({ N }_{ o }{ e }^{ -\lambda t }\)
When t = T_α, then \(N={ N }_{ o }{ e }^{ -\lambda \times \frac { 1 }{ \lambda } }={ N }_{ o }{ e }^{ -1 }=\frac { 1 }{ e } { N }_{ o }\)
or \(\frac { N }{ { N }_{ o } } =\frac { 1 }{ e } =\frac { 1 }{ 2.718 } =0.368\)
So, the fraction of the element left behind is 0.368. The fraction of the element which disappear = 1 - 0.368 = 0.368.

Decay constant \(\lambda =\frac { 0.6931 }{ T } =\frac { 0.6931 }{ 24.1 } \) day^-1
Also, \(\frac { N }{ { N }_{ o } } =\frac { 10 }{ 100 } =\frac { 1 }{ 10 } \)
Now, \(\frac { N }{ { N }_{ o } } ={ e }^{ -\lambda t };\frac { 1 }{ 10 } ={ e }^{ -\lambda t };{ e }^{ -\lambda t }=10\)
Taking logs, we get λt = log, 10 = 2.3
t = \(t=\frac { 23 }{ \lambda } =\frac { 2.3\times 24.1 }{ 06931 } \) days = 19.97 days

(i) Number of atoms present initially
\(=\frac { 6.023\times { 10 }^{ 23 }\times 2.2\times { 10 }^{ -3 } }{ 11 } =1.2\times { 10 }^{ 20 }\)
(ii) Number of atoms present in 5 μg of the sample, N
\(=\frac { 6.023\times { 10 }^{ 23 }\times 5\times { 10 }^{ -6 } }{ 11 } =2.74\times { 10 }^{ 17 }\)
The activity of the sample = λ\(N=\frac { 0.693 }{ { T }_{ 1/2 } } \times N\)
\(=\frac { 0.693 }{ 1224 } \times 2.74\times { 10 }^{ 17 }\)
= 1.55 x 10¹⁴ disintegrations/second