How to Calculate the decay constant, half-life and the mean life for a radioisotope which activity is found to decrease by 25% in one week??

1 Answer
May 21, 2018

#lambda~~0.288color(white)(l)"week"^(-1)#
#t_(1/2)~~2.41color(white)(l)"weeks"#
#tau~~3.48color(white)(l)"weeks"#

Explanation:

The first-order decay constant #lambda# comprises the expression for the decay activity at a particular time #A(t)#.

#A(t)=A_0*e^(-lambda*t)#
#e^(-lambda*t)=(A(t))/A_0=1/2#

Where #A_0# the activity at time zero. The question suggests that #A(1color(white)(l)"week")=(1-25%)*A_0#, thus

#e^(-lambda*1color(white)(l)"week")=(A(1color(white)(l)"week"))/(A_0)=0.75#

Solve for #lambda#:

#lambda=-ln(3/4)/(1color(white)(l)"week")~~0.288color(white)(l)"week"^(-1)#

By the (self-explanatory) definition of decay half-life

#e^(-lambda*t_(1/2))=(A(t_(1/2)))/A_0=1/2#
#-lambda*t_(1/2)=ln(1/2)#
#t_(1/2)=ln2/(lambda)~~2.41color(white)(l)"weeks"#

Mean life #tau# represents the arithmetic mean of all individual lifetimes and is equal to the reciprocal of the decay constant.

#tau=1/lambda=3.48color(white)(l)"weeks"#