Radioactive material equation

[Sabi]

Hi guys,

I could do with some guidance on this question and if I'm headed in the right direction? I (think) I'm ok with part a/b but correct me if I'm wrong thinking that. Part c however has me confused and this is all I have come to.

Many thanks for any help

Sab

 

[skeeter]

[MATH]\dfrac{m}{m_0} = e^{\lambda t} \implies \dfrac{m_0}{m} = e^{-\lambda t}[/MATH]
[MATH]\ln\left(\dfrac{m_0}{m} \right) = -\lambda t[/MATH]
try (c) again ...

 

[Sabi]

[MATH]\dfrac{m}{m_0} = e^{\lambda t} \implies \dfrac{m_0}{m} = e^{-\lambda t}[/MATH]
[MATH]\ln\left(\dfrac{m_0}{m} \right) = -\lambda t[/MATH]
try (c) again ...

Don't really know what to do to be honest. I'm sure its straight forward I guessed that the half life should be half the mass which is why I wrote what I did.

Is it just Mh = lamba-1/2?

 

[skeeter]

[MATH]m = \dfrac{m_0}{2} \implies \ln\left(\dfrac{m_0}{m}\right) = \ln(2)[/MATH]
therefore ...

[MATH]\ln(2) = -\lambda t_h \implies t_h = \, ?[/MATH]

 

[Sabi]

[MATH]m = \dfrac{m_0}{2} \implies \ln\left(\dfrac{m_0}{m}\right) = \ln(2)[/MATH]
therefore ...

[MATH]\ln(2) = -\lambda t_h \implies t_h = \, ?[/MATH]

Ln(2) / - lambda = th

 

[Sabi]

Still going round in circles with getting a grip on this. Can anyone give pointers on my previous reply.

 

[skeeter]

Still going round in circles with getting a grip on this. Can anyone give pointers on my previous reply.

why the need for affirmation? ... are you not confident in your basic algebra skills?

[MATH]t_h = -\dfrac{\ln(2)}{\lambda}[/MATH]

 

[Sabi]

why the need for affirmation? ... are you not confident in your basic algebra skills?

[MATH]t_h = -\dfrac{\ln(2)}{\lambda}[/MATH]

I see... Was focusing on (Mh) when it is Th....?‍ doh.. Thank you for your replies.