Exactly 1/16 of a given amount of protactinium-234 remains after 26.76 hours. what is the half-life of protactinium-234?
How do i carry out this problem? [details please]
*PLEASE HELP*
2006-11-02 10:33:58 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
It is very simple. This method will help you when the fraction is not a nice one like 1/16.
At time zero you have, let's say N0 = 1600 atoms of protactinium-234. After 26.76 hours, 1/16 remain:
Nt = 1600 / 16 = 100.
Now in radioactive decay:
Nt = N0 * exp (-Lamda * t).
You have t, Nt, N0. Calculate lambda
100 = 1600 exp (-Lambda * 26.76)
0.0625 = exp (-Lambda * 26.76)
Taking Ln on both sides
Ln 0.0625 = - Lambda * 26.76
Lambda = - (Ln 0.0625) / 26.76 = 0.10361 h^(-1)
Half life = (Ln 2) / Lambda = 6.69 hours
2006-11-02 11:34:19 · answer #1 · answered by Dr. J. 6 · 0⤊ 0⤋
If x is the half-life of protactinium-234, then:
In x hours remains 1/2 of the initial mass
In 2x hours remains 1/4 of the initial mass
In 3x hours remains 1/8 of the initial mass
In 4x hours remains 1/16 of the initial mass
So 4x = 26.76, x = 26.76/4, x = 6.69 hours
That's the protactinium-234 half-life.
2006-11-02 10:50:37 · answer #2 · answered by Dimos F 4 · 0⤊ 0⤋
Half life is the time it takes half of a sample to decay.
1/2 1/4 1/8 1/16
In the first half life 1/2 decays, 1/ .2 of that is 1/4, 1/2 of that is 1/8 ,1/2 of that is 1/16 That is 4 half lifes.
26.76 /4 =6.69hours
2006-11-02 11:11:04 · answer #3 · answered by science teacher 7 · 0⤊ 0⤋