Question

b) Uranium disintegrates at a rate proportional to the amount present at any instant.
If my and m2 grams of uranium are present at times T1 and T2 respectively show that the half
mi
life of Uranium is (T2 -T1) log 2/log
m2

Answer 1

not understand please say one more time

Answer 2

Answer:

Hence Half Life = T = (T2-T1)ln2 / (ln(m1/m2)

Step-by-step explanation:

We know that,

$\frac{m}{m_{0} } = 2^{\frac{-t}{T} } \\\\log_{2}\frac{m}{m_{0} } = \frac{-t}{T} \\\\T = \frac{tln2}{ln\frac{m_{0} }{m} }\\\\lnm_{0} - lnm= \frac{tln2}{T} \\ So,\\ lnm_{0} = \frac{tln2}{T} + lnm = constant \\\\Now, we have ,\\ \frac{tln2}{T} + lnm = constant$

So for m1 we have

$\frac{t_{1} ln2}{T} + lnm_{1} = constant$

And for m2 we have

$\frac{t_{2} ln2}{T} + lnm_{2} = constant$

Equating them we get,

$\frac{t_{1} ln2}{T} } + lnm_{1} = \frac{t_{2} ln2}{T} + lnm_{2}\\\\\\lnm_{2} -lnm_{1} = \frac{(t_{1} -t_{2} )ln2 }{T}\\ \\Half life = T = \frac{(t_{1} -t_{2} )ln2 }{ln\frac{m_{2} }{m_{1} } } \\\\\\Half life = T = \frac{(t_{2} -t_{1} )ln2 }{ln\frac{m_{1} }{m_{2} } }$