Introduction to Original half life :-
The formula for Exponential decay is N = N0eKt
N0 is the initial population and the decay rate is k
N is the final population after the time t with the decay rate k.
Half- Life :-
In half life `N = N_0/2 `
Plug in the the value of N as `N_0/2` in the initial equation we get .
`N_0/2``=N_0 e^(kt)`
Now canceling N_0 in both side and solving we get .
`1/2 = e^(kt)`
`2^-1 = e^(kt)`
Take ln on both sides of above equation we get
` ln (2^-1) = ln (e^(kt))`
Solving it we get
-1 ln(2) = kt.
Solving for t we get
` t = -ln2/k`
Half life time ` (t) = -ln(2)/k`
Now lets see some solved problems in the topic origianl half life.
Original Half Life Solved Example Problem
The original half-life of an plutonium particle is 18,000 years. If 30 gram are present now, find how long it will take until only 20% of the original atomic particle remains.
Solution To Half life Day One Solved Example Probblem 1:-
The half period time is 18,000 years.
The initial amount is 30 gram..
We need to find the time it will take to make the 20% of original plutonium particle.
First we need to find the decay constant k.
The formula used for decay constant is
Plug in the values in the formula we get.
`18, 000 = - ln (2)/ k`
Now arraigning the above expression to find the value k we get
`k = - ln (2)/ (18,000)`
The value of ln(2) is 0.6931.
`= -0.6931 /( 18,000)`
`(-0.6931) / (18 000) = -0.0000385`
Therefore the value of decay constant is -0.0000385.
Now we have to find the time when there will be only 20% of atomic particle remains.
`6 = 30 xx e^(-0.0000385 t)`
Now divide by 30 on both side of the above expression.
`6/ 30 = (30 xx e^(-0.0000385 t))/ 30`
`0.2 = e^(0.0000385 t)`
Taking ln on both side of the above expression we get
`ln (0.2) = ln (e^(-0.0000385 t))`
`ln (e^(-0.0000385 t))` can be written as `-0.0000385 t.`
`ln(0.2) = -0.0000385 t..`
Now divide by `- 0.0000385` on both side of the above expression we get .
`ln(0.1)/ -0.0000385 = (-0.0000385 t)/ -0.0000385.`
By solving the above expression we get
`ln(0.2)/ -0.0000385 = t`
By solving the above fraction we get
41,803 = t
20 percent of 30 gram plutonium particle remains after 41,803 years.
Original Half Life Practice Problem
The half-life of an Barium particle is 16,000 years. If 40 gram are present now, find how long it will take until only 10% of the original atomic particle remains.
Answer:-
10 percent of 40 gram plutonium particle remains after 40,859 years.
The formula for Exponential decay is N = N0eKt
N0 is the initial population and the decay rate is k
N is the final population after the time t with the decay rate k.
Half- Life :-
In half life `N = N_0/2 `
Plug in the the value of N as `N_0/2` in the initial equation we get .
`N_0/2``=N_0 e^(kt)`
Now canceling N_0 in both side and solving we get .
`1/2 = e^(kt)`
`2^-1 = e^(kt)`
Take ln on both sides of above equation we get
` ln (2^-1) = ln (e^(kt))`
Solving it we get
-1 ln(2) = kt.
Solving for t we get
` t = -ln2/k`
Half life time ` (t) = -ln(2)/k`
Now lets see some solved problems in the topic origianl half life.
Original Half Life Solved Example Problem
The original half-life of an plutonium particle is 18,000 years. If 30 gram are present now, find how long it will take until only 20% of the original atomic particle remains.
Solution To Half life Day One Solved Example Probblem 1:-
The half period time is 18,000 years.
The initial amount is 30 gram..
We need to find the time it will take to make the 20% of original plutonium particle.
First we need to find the decay constant k.
The formula used for decay constant is
Plug in the values in the formula we get.
`18, 000 = - ln (2)/ k`
Now arraigning the above expression to find the value k we get
`k = - ln (2)/ (18,000)`
The value of ln(2) is 0.6931.
`= -0.6931 /( 18,000)`
`(-0.6931) / (18 000) = -0.0000385`
Therefore the value of decay constant is -0.0000385.
Now we have to find the time when there will be only 20% of atomic particle remains.
`6 = 30 xx e^(-0.0000385 t)`
Now divide by 30 on both side of the above expression.
`6/ 30 = (30 xx e^(-0.0000385 t))/ 30`
`0.2 = e^(0.0000385 t)`
Taking ln on both side of the above expression we get
`ln (0.2) = ln (e^(-0.0000385 t))`
`ln (e^(-0.0000385 t))` can be written as `-0.0000385 t.`
`ln(0.2) = -0.0000385 t..`
Now divide by `- 0.0000385` on both side of the above expression we get .
`ln(0.1)/ -0.0000385 = (-0.0000385 t)/ -0.0000385.`
By solving the above expression we get
`ln(0.2)/ -0.0000385 = t`
By solving the above fraction we get
41,803 = t
20 percent of 30 gram plutonium particle remains after 41,803 years.
Original Half Life Practice Problem
The half-life of an Barium particle is 16,000 years. If 40 gram are present now, find how long it will take until only 10% of the original atomic particle remains.
Answer:-
10 percent of 40 gram plutonium particle remains after 40,859 years.
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