The nucleus 115/48cd, after two successive β?decay will give

 

According to Nuclear activity, we can write

$N_0\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{2}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{4}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{8}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{16}=\left(0.0625\right)N_0$  

Time required = 4 ×  $t_{\frac{1}{2}}=20yrs$

From Radioactive Decay Law,
dN/dt = λ? N + λ? N = λ_eff N
⇒ λ_eff = λ? + λ? ⇒ ln (2)/T = ln (2)/T? + ln (2)/T? ⇒ T = (T? ) / (T? + T? )
(where T, T? , and T? are half-lives)

The law of radioactive decay is N = N? e? λt, where N is the amount remaining at time t.
Given that at time t, N/N? = 9/16.
So, 9/16 = e? λt
At time t/2, the fraction remaining will be N'/N?
N' = N? e? λ ( t/2 ) = N? (e? λt)¹/²
Substituting the value of e? λt:
N' = N? (9/16)¹/² = N? (3/4)
The fraction remaining is N'/N? = 3/4.

$A=A_0{e}^{-\lambda t}$

$\begin{array}{rr}& 500=700{\mathrm{e}}^{-\lambda \mathrm{t}}\\ & \lambda \mathrm{t}=\mathrm{l}\mathrm{n}?\frac{7}{5}\\ & \frac{\mathrm{l}\mathrm{n}?2}{{\mathrm{t}}_{1/2}}\times 30=\mathrm{l}\mathrm{n}?\frac{7}{5}\\ & ?\mathrm{}{\mathrm{t}}_{1/2}=\frac{\mathrm{l}\mathrm{n}?2\times 30}{\mathrm{l}\mathrm{n}?\frac{7}{5}}\end{array}$

${\mathrm{t}}_{1/2}=61.8\mathrm{m}\mathrm{i}\mathrm{n}\approx 62\mathrm{m}\mathrm{i}\mathrm{n}$ .

t? /? = 3 days = 72 hours
dN/dt = λN = (ln2/t? /? ) N
= (0.693 × 6.02×10²³ × 2×10? ³) / (72 × 3600 × 198)
= 1.618 × 10¹³
= 16.18 × 10¹² disintegration/second