Question

The radioactive substance krypton-91 has a very short half-life of about 10 seconds. What is the decay constant for this substance?

Answer 1

\[\frac{1}{2}=e^{10k}\] solve for \(k\)

Answer 2

where do i go from here?

Answer 3

take the log, divide by 10

Answer 4

divide 1/2 by 10?

Answer 5

oh no
you need to get the variable out of the exponent
first step is to take the log, i.e. write
\[\frac{1}{2}=e^{10k}\] in equivalent exponential form as
\[\ln(\frac{1}{2})=10k\] then divide by 10 to get \(k\)

Answer 6

i meant equivalent logarithmic form

Answer 7

i got -0.06931471805 @satellite73 i have a feeling this is wrong

Answer 8

^ that answer is correct for k.

Answer 9

\[A = A _{0} e^{kt}\]A is amount remaining, Ao is initial amount. The half life is when half the initial amount remains, or 1/2 Ao
so for half life:
\[\frac{ 1 }{ 2}A _{0} = A _{0}e ^{kt}\]
divide it all by Ao:
\[\frac{ 1 }{ 2 } = e ^{kt}\]
Then enter t = 10seconds, and solve by taking ln of both sides as satellite showed.

Answer 10

To check your k value, enter it into the equation, and enter t=10 seconds, and check your equation gives you 1/2Ao:
\[A = A _{0} e^{-0.06931*10} \]
Which is:
\[A = A _{0}*0.5\]

Answer 11

thank you so much!! so the final answer would be -0.06931471805?

Answer 12

Which means that after 10 seconds, the amount remaining is half the initial amount, which means 10 seconds is the half life, as stated in the original question "The radioactive substance krypton-91 has a very short half-life of about 10 seconds."

Answer 13

Yep, though you can probably round it to four or five decimal places. The decay constant k is -0.06931

Answer 14

thanks so much!! other sources have told me it is a positive number though

Answer 15

0.06931471805 does that make sense?

Answer 16

A positive k value would mean exponential growth, not decay.

Answer 17

Try it, enter a positive value into
\[A = A _{0} e^{0.06931*10} \]
\[A = A _{0}*2\]
which means the amount of krypton remaining after 10 seconds is actually double the initial amount...

Answer 18

Oh i guess k would be positive if you used the formula
\[A = A _{0} e^{-kt}\]

Answer 19

great, thank you so

Answer 20

so much!

Answer 21

Oh and i guess you should have units of 1/s or
\[k = 0.06931 s ^{-1}\]
ie that's how much decays per second.

Answer 22

thanks :)

Answer 23

@agent0smith can you possibly help me with one more question i just posted? you were very helpful with this one!