The half life of a radioactive element is 30 seconds. In what period of time would the activity of the sample be reduced to one - sixteenth of the original activity?

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Answer 1 · 2018-02-17T17:36:47

Working formula: #A=A_0 * e^-(kt)#

You can simplify the above working formula to the following form:

#kt=ln(A_0/A)#

where #A_0# is the activity of the element at time #t=0# and #A# is the activity at a given time.

Now, we know that the formula for half-life (which can be obtained from the above equation itself) is given by:

#t_("1/2")=(ln 2)/k#.

So, here the half-life is #30# seconds. Using this information, we find the decay constant #k#.

#k=ln2/"30 s"#

Now, using this value of #k#, we get the time at which the activity of the element becomes #1/16# of original activity, i.e

#A=A_0/16#

So

#t=1/(ln 2/"30 s")*ln(A_0/(A_0/16))#

# t=("30 s"/ln2)*ln (2^4)#

# t=("30 s"/ln2)*4 ln2#

# t="30 s" * 4#

# t= "120 s"#

Therefore, the activity of the radioactive element becomes one-sixteenth of its original activity at #120# seconds.