Decayed = 60 %
Left undecayed = 40 %(ie) \(\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{40}{100} \)
\(\mathrm{~T}_{\frac{1}{2}}=3.8 \text { days } \)
\(\mathbf{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t} \)
\(\frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda t} \)
\(\frac{40}{100}=\mathrm{e}^{-\lambda t} \Rightarrow \frac{100}{40}=2.5=\mathrm{e}^{\lambda t} \)
\(\therefore \mathrm{e}^{\lambda t} \) = 2.5
Taking log on both sides
\(\lambda t=\ln [2.5]=2.3026 \times \log (2.5)=2.3026 \times 0.3974 \)
\(t=\frac{0.9163}{\lambda}=\frac{0.9163}{0.6931} \times T_{1 / 2} \)
\(t=1.322 \times 3.8=5.022 \text { days } \)

For A, Half life of  \(, \mathrm{T}_{\mathrm{A}} \) = 20 minutes, n = 4
For B, Half life of  \(, T_{B}\) = 40 minutes, n = 2
\(\mathrm{N}_{01}=\mathrm{N}_{02}=\mathrm{N}_{0}\)
sample left A, \(\frac{N_{1}}{N_{0}}=\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{4}=\frac{1}{16}\)
A-sample decayed \(=1-\frac{N_{1}}{N_{0}}=1-\frac{1}{16}=\frac{15}{16}\)
sample Ieft B, \(\frac{N_{1}}{N_{0}}=\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}\)
B-sample decayed \(=1-\frac{N_{2}}{N_{0}}=1-\frac{1}{4}=\frac{3}{4}\)
Ratio of decayed number of A and B \(=\frac{\frac{15}{16}}{\frac{1}{4}}=\frac{15}{16} \times \frac{4}{3}=\frac{5}{4}\)
Ratio of decayed number of A and |B = 5:4

Density \(\rho=2.3 \times 10^{17} \mathrm{kgm}^{-3}\), Mass M = 5.97 x 10²⁴ kg
\(\rho=\frac{M}{V}=\frac{M}{\frac{4}{3}\pi R^3}\)
\(R=\left[\frac{M}{\frac{4}{3} \pi \rho}\right]^{\frac{1}{3}}=\left[\frac{3 \mathrm{M}}{4 \pi \rho}\right]^{\frac{1}{3}}=\left[\frac{3 \times 5.97 \times 10^{24}}{4 \times 3.14 \times 2.3 \times 10^{17}}\right]^{\frac{1}{3}}=\left[0.62 \times 10^{7}\right]^{\frac{1}{3}}\)
R = 183.7 m
R ≈180 m

Binding energy BE = [Zm_H + Nm_n - M_A]c²
For helium nucleus, Z = 2, N = A–Z = 4–2 = 2
Mass defect
Δm = [(2 x 1.00785u) + (2 x 1.008665 u) -4.00260u] Δm = 0.03043 u
B.E = 0.03043u x c²
B.E = 0.03043 x 931 MeV = 28.33 MeV
[∵ luc² = 931 MeV]
The binding energy of the \(_{ 2 }^{ 4 }{ He }\) nucleus is 28.33 MeV.

\(T_{1 / 2}=1600 \text { years }=1600 \times 365 \times 24 \times 60 \times 60 s\)
R = 1 curie = 3.7 x 10¹⁰ Bq, Show that m = 1g
R = λN
Number of atoms Present, N = \(\frac{\mathrm{R}}{\lambda}=\frac{\mathrm{R}}{0.6931} \mathrm{~T}_{1 / 2}\)
Mass of 6.023 x 10²³ atoms of  \({ }_{88}^{226} R a=226 g\)
Mass of 1 atom of  \({ }_{88}^{226} \mathrm{Ra}=\frac{226}{6.023 \times 10^{23}} \mathrm{~g}\)
Mass of N atoms of \({ }_{88}^{{ }{266}} \mathrm{Ra}=\frac{226}{6.023 \times 10^{23}} \times \mathrm{Ng}\)
Mass of N atoms of \({ }_{88}^{226} \mathrm{Ra}(\mathrm{m})=\frac{226}{6.023 \times 10^{23}} \times \frac{\mathrm{R}}{0.6931} \mathrm{~T}_{1 / 2} \mathrm{~g}\)
\(\mathrm{m}=\frac{226}{6.023 \times 10^{23}} \times \frac{3.7 \times 10^{10}}{0.6931} \times 1600 \times 365 \times 24 \times 60 \times 60 \mathrm{~g}\)
m = 1.01 g