Question #1dc8f

1 Answer
May 1, 2016

Answer:

It is four.

Explanation:

The half life of any substance is given by:
#t_(1/2)=(In2)/lamda#

Hence

#t_(1/2(1))=(In2)/lamda_1# .... and.... #t_(1/2(2))=(In2)/lamda_2#

We want to find:

#Ratio=t_(1/2(1))/(t_(1/2(2)))#

Substituting the previous equations into the above gives
#Ratio=t_(1/2(1))/(t_(1/2(2)))=lamda_2/lamda_1# [equation A]

For a sample of N atoms, the activity, A, is given by:

#A=lamda*N#

Where #lamda# is the decay constant.

So for sample 1: #A_1=lamda_1*N_1# [equation 1]

and for sample 2: #A_2=lamda_2*N_2# [equation 2]

We are told that that #N_1=2N_2# or#N_1/2=N_2#
and also #A_2=2A_1#

Substituting both of these into equation 2 we get:
#2A_1=lamda_2*N_1/2#
So
#A_1=lamda_2/4*N_1# [equation 3]

Comparing equation 1 and 3 we can see that:

#lamda_1=lamda_2/4#

and so #lamda_2/lamda_1=4#

Hence the ratio of halve lives is 4 (by comparing this result to equation A)