Question 289: Atoms & Nuclei

Question

A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 $$ 	imes $$ 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive materail B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :

Accepted Answer

Let number of nuclei present in material A is N
A
and in meterial  B  is N
B
According  to  question,
N
A
= 2N
B
We know, activity (A) = $$\lambda $$N.
$$ 	herefore $$   $$\lambda $$
A
N
A
= 10 . . . . . . (1)
and $$\lambda $$
B
N
B
= 20 . . . . . .
 . (2)
$$ 	herefore $$   By dividing (1) by (2), we get
$${{{\lambda _A}{N_A}} \over {{\lambda _B}{N_B}}}$$ = $${1 \over 2}$$
$$ \Rightarrow $$   $${{{\lambda _A}} \over {{\lambda _B}}} 	imes 2$$ = $${1 \over 2}$$ [as  N
A
= 2N
B
]
$$ \Rightarrow $$    $${{{\lambda _A}} \over {{\lambda _B}}}$$ = $${1 \over 4}$$
We know,
T$$_{{1 \over 2}}$$  $$ \propto $$ $${1 \over \lambda }$$
$$ 	herefore $$   $${{{{\left( {{T_{{1 \over 2}}}} ight)}_A}} \over {{{\left( {{T_{{1 \over 2}}}} ight)}_B}}}$$ = $${{{\lambda _B}} \over {{\lambda _A}}} = 4$$
$$ \Rightarrow $$   $${\left( {{T_{{1 \over 2}}}} ight)_A}$$ = 4$${\left( {{T_{{1 \over 2}}}} ight)_B}$$
$$ 	herefore $$   By checking options you can see  possible  value of $${\left( {{T_{{1 \over 2}}}} ight)_A}$$ = 20 days and $${\left( {{T_{{1 \over 2}}}} ight)_B}$$ = 5 days

Answer

5 days and 10 days

Answer

10 days and 40 days

Answer

20 days and 5 days

Answer

20 days and 10 days

Atoms & Nuclei

A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 $ \times $ 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive materail B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :

Choose the correct answer: