Question

226Ra has a half-life of 1599 years.
How much is left after 1000 years if the initial amount was 10 g?

Answer 1

i dont know sorry

Answer 2

\[y=ae^{bx}\]\[x=1599\]\[a=10g\]\[y=?\]

Answer 3

Meh @anabelle

Answer 4

sos

Answer 5

lol

Answer 6

XD

Answer 7

I did it like this:
(0,10)\[0=10e^{b \times1599}\]\[0=e^{1599b}\]

Answer 8

... aaaand now I'm stuck

Answer 9

\[[\text{Ra}](t)=[\text{Ra}]_0 (\tfrac12)^{t/t_{1/2}}\]

Answer 10

Okay?

Answer 11

you have the half life so you should the half life formula not exponential decay formula
the difference is that the base is 1/2 not e

Answer 12

t is the time (you can use x if you really want to)

t_1/2 is the half life

[Ra](t) is the concentration of Radon at time t

[Ra]_0 is the initial concentration of Radon

Answer 13

o.o
We did it like this in class:\[y=ae^{bx}\]\[\frac{1}{2}=ae^{b\times 0}\]\[a=\frac{1}{2}\]\[y=\frac{1}{2}e^{bx}\]Is that what you were referring to?

Answer 14

If yes, /headdesk I was stupid and didn't realize LOL

Answer 15

and I meant to type this \[\frac{1}{2}=ae^{bx}\]

Answer 16

its kinda similar

Answer 17

\[[\text{Ra}](1000)=10[\text g] (\tfrac12)^{1000/1599}\]

Answer 18

Uh, can we simplify into easier-to-read variables?

Answer 19

you can if you wish, but you should have 1/2 as the base not e

Answer 20

\[\large y(x)=a(\tfrac12)^{x/x_{1/2}}\]

Answer 21

Ah, okay... what is x 1/2? Is that x (while x is b)?

Answer 22

just add by tens and if it is enough then you get the answer, from 10 - 100 it is not that hard =)

Answer 23

t_1/2=x_1/2= the half life

Answer 24

@carolinagb02 I'm sure these formulas exist for a reason

Answer 25

@UnkleRhaukus so\[y=a\frac{1}{2}^{bx}\]or am I wrong?

Answer 26

it depends how you define b

Answer 27

b is the rate of decay

Answer 28

i suppose you can have e as the base

if

\[b=\frac{-\ln(1/2)}{t_{1/2}}=\frac{\ln(2)}{t_{1/2}}\]

Answer 29

wat

Answer 30

Maybe I should just go over this in class and get back to you on this lol

Answer 31

It's 6:45 AM right now... class starts at 7:30

Answer 32

I still have to sleep T^T

Answer 33

if you really wanted to use that first formula

\[y=ae^{bx}\]

you find \(b\) by \(b=\ln2/\text{half life}\)

Answer 34

I have to go, but I'll keep that in mind

Answer 35

x is 1000

Answer 36

Actually, we do it like this:
\[y=ae^{bx}\]\[1=2e^{1599(x)}\]\[\frac{1}{2}=e^{1599x}\]\[ln\frac{1}{2}=1599x\]\[x\approx. -4.3349\]

Answer 37

\[\large y(t)=y_0e^{-\ln 2 \frac{t}{t_{1/2}}}\\
\\
=10e^{-\ln 2\frac{1000}{1599}}\\
\approx 6.5\]

Answer 38

wat o_o