The problem says: It takes 300,000 years for a certain radioactive substance to decay to 30% of its original amount. What is its half-life?

The result is 173,000 years, but I don't see how it is obtained. I tried solving for x in f(x)=300,000e^(300,000×x)=0.3, which is approximately x=0.00004. Then I solved for x in f(x)=e^(-0.00004×x)=1/2, which is approximately x=173286. That's the closest I got to the actual result. What am I doing wrong? How do I obtain 173,000? Or maybe my answer is correct and 173,000 it's just an approximation? Thanks.

 

 

4 hours ago, KFS said:

The problem says: It takes 300,000 years for a certain radioactive substance to decay to 30% of its original amount. What is its half-life?

The result is 173,000 years, but I don't see how it is obtained. I tried solving for x in f(x)=300,000e^(300,000×x)=0.3[/quote]

Frankly, it looks like you are trying to apply a formula without understand the formula or thinking about how it applies to this problem!  First, that "300,000" multiplying the exponential makes no sense!  What you have there would give f(0)= 300,000.   What would that 300,000 even mean?  f(x)  should give an amount either in some mass units or as a percent or fraction.  And the number multiplying the exponential is the same.  Since you have "= 0.3", a decimal fraction, the number multiplying the exponential must be 1.00.  \EX] will give the half life when [tex]e^{\alpha  x}= 0.5[/tex] so when  [tex]\alpha = 0.5/x

 

 

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, which is approximately x=0.00004. Then I solved for x in f(x)=e^(-0.00004×x)=1/2, which is approximately x=173286. That's the closest I got to the actual result. What am I doing wrong? How do I obtain 173,000? Or maybe my answer is correct and 173,000 it's just an approximation? Thanks.

 

8 minutes ago, HallsofIvy said:

 

 

 

How do I do it then? Can you explain?

4 hours ago, KFS said:

The problem says: It takes 300,000 years for a certain radioactive substance to decay to 30% of its original amount. What is its half-life?

The result is 173,000 years, but I don't see how it is obtained. I tried solving for x in f(x)=300,000e^(300,000×x)=0.3, which is approximately x=0.00004. Then I solved for x in f(x)=e^(-0.00004×x)=1/2, which is approximately x=173286. That's the closest I got to the actual result. What am I doing wrong? How do I obtain 173,000? Or maybe my answer is correct and 173,000 it's just an approximation? Thanks.

You might want to approach this differently.  How many half lives would it take to get to 30%?  If you knew how many half lives, you would then know how many half lives equals 300,000 years.

14 minutes ago, OldChemE said:

You might want to approach this differently.  How many half lives would it take to get to 30%?  If you knew how many half lives, you would then know how many half lives equals 300,000 years.

The formula is f'=-(kappa)×f, where f=f(0)e^(-kappa×t), or (1/kappa)×ln2. In this case what would each term mean? I didn't understand that.

Edited October 2, 20205 yr by KFS

4 hours ago, KFS said:

The problem says: It takes 300,000 years for a certain radioactive substance to decay to 30% of its original amount. What is its half-life?

The result is 173,000 years, but I don't see how it is obtained. I tried solving for x in f(x)=300,000e^(300,000×x)=0.3, which is approximately x=0.00004. Then I solved for x in f(x)=e^(-0.00004×x)=1/2, which is approximately x=173286. That's the closest I got to the actual result. What am I doing wrong? How do I obtain 173,000? Or maybe my answer is correct and 173,000 it's just an approximation? Thanks.

The formula N =N₀ e^-ct

Gives you  the number of remaining atoms(N) starting from an Original number of atoms(N₀) when you know the time of decay(t) and the decay constant (c)

You are given t and N/N_0.

From this, you should be able to solve for c

Further, cT = ln(2),  where T is the half-life.

This should be enough to solve the problem (though as you suspected, 173,000 yrs is a "rounded out" answer.)

 

4500,000 years will reduce to 30% with e^{500,000*\alpha}= 0.3.  500,000*\alpha= ln(0.3) so \alpha= \frac{ln(0.3)}{500,000}= ln\left(3^{1/500000}\right).  And then to find the half life, set e^{\alpha x}= e^{ln\left(3^{1/500000}\right)t}= 3^{(1/500000)t= 0.5 and solve for t.  

Edited October 3, 20205 yr by HallsofIvy

1 hour ago, HallsofIvy said:

4500,000 years will reduce to 30% with e^{500,000*\alpha}= 0.3.  500,000*\alpha= ln(0.3) so \alpha= \frac{ln(0.3)}{500,000}= ln\left(3^{1/500000}\right).  And then to find the half life, set e^{\alpha x}= e^{ln\left(3^{1/500000}\right)t}= 3^{(1/500000)t= 0.5 and solve for t.  

How did you get the 500,000?

Is this correct?

11 hours ago, KFS said:

Is this correct?

It's the answer I get.