Question

Half-lives of two radioactive elements A and B are
20 minutes and 40 minutes, respectively. Initially,
the samples have equal number of nuclei. After
80 minutes , the ratio of decayed numbers of A and
B nuclei will be :-
(1) 5 : 4 (2) 1 : 16
(3) 4 : 1 (4) 1 : 4

Answer 1

As we learnt in

Number of nuclei in terms of half life -

N=\frac{N_{0}}{2^{t/t_{1/2}}}

- wherein

Very useful to determine number of nuclei in terms of half life

A  (T1/2 = 20 minutes)             B  (T1/2 = 40 minutes)

N_{A}=\frac{N_{0}}{2^{t/T_{1/2}}}N_{B}=\frac{N_{0}}{2^{t/t_{1/2}}

at t = 80 min. N_{A}=\frac       {N_{0}}{2^{4}}=\frac{N_{0}}{16} at t = 80 min.

N_{B}=\frac{N_{0}}{2^{2}}=\frac{N_{0}}{4}

Decayed number of  A=\frac{N_{0}-N_{0}}{16}=\frac{15}{16}N_{0}

Decayed number of B=\frac{3N_{0}}{4}

Ratio =\frac{\frac{15}{16}N_{0}}{\frac{3}{4}N_{0}}=\frac{5}{4}Correct option is 4

Option 1)       1 : 16    This is an incorrect option.

Option 2)      4 : 1      This is an incorrect option.

Option 3)      1 : 4      This is an incorrect option.

Option 4)      5 : 4      This is the correct option.

Answer 2

Answer: 5:4

Explanation:

Given that half lives of two radioactive elements A and B are 20 min and 40 minutes respectively.

We know that,

$\text{Half-Life}(T_{\frac{1}{2}})=\frac{\ln2}{\lambda}$

Applying this formula for for A and B

We have,

$\text{Half-Life of A}=\frac{\ln2}{\lambda_{A}}\\\;\\20=\frac{\ln2}{\lambda_{A}}\\\;\\\lambda_A=\frac{\ln2}{20}$

And similarly,

$\lambda_B=\frac{\ln2}{40}$

Now, Let the nuclei of A and B  initially be  $N_0$

and after 80 minutes it becomes $N_A$ and $N_B$

We know the relation between Nuclei at time t  is given as,

$N=N_0e^{-\lambda t}$

Using this formula for A and B,

We have,

$N_A=N_0e^{\lambda_{A}\times80}\\\;\\N_A=N_0e^{-\frac{\ln2}{20}\times80}\\\;\\N_A=N_0e^{-4\ln2}\\\;\\N_A=N_0e^{-\ln2^4}\\\;\\N_A=N_0e^{\ln\frac{1}{2^4}}\\\;\\N_A=\frac{N_0}{16}$

Similarly for B,

$N_B=N_0e^{\lambda_{B}\times80}\\\;\\N_B=N_0e^{-\frac{\ln2}{40}\times80}\\\;\\N_B=N_0e^{-2\ln2}\\\;\\N_B=N_0e^{-\ln2^2}\\\;\\N_B=N_0e^{\ln\frac{1}{2^2}}\\\;\\N_B=\frac{N_0}{4}$

Now,

$\text{Decayed Nuclei of A}=N_0\;-\;N_A\\\;\\=N_0-\frac{N_0}{16}\\\;\\=\frac{15N_0}{16}$

and,

$\text{Decayed Nuclei of B}=N_0\;-\;N_B\\\;\\=N_0-\frac{N_0}{4}\\\;\\=\frac{3N_0}{4}$

Therefore,

$\frac{\text{Decayed Nuclei of A}}{\text{Decayed Nuclei of B}}=\frac{\frac{15N_0}{16}}{\frac{3N_0}{4}}\\\;\\=\frac{15\times4}{16\times3}\\\;\\=\frac{5}{4}$

So the ratio of decayed  nuclei of A and B is 5:4