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Lorico [155]
3 years ago
7

What traction of the radioisotoperemains in the body after one day?​

Physics
1 answer:
r-ruslan [8.4K]3 years ago
5 0

The fraction of radioisotope left after 1 day is (\frac{1}{2})^{\frac{1}{\tau}}, with the half-life expressed in days

Explanation:

The question is incomplete: however, we can still answer as follows.

The mass of a radioactive sample after a time t is given by the equation:

m(t)=m_0 (\frac{1}{2})^{\frac{t}{\tau}}

where:

m_0 is the mass of the radioactive sample at t = 0

\tau is the half-life of the sample

This means that the mass of the sample halves after one half-life.

We can rewrite the equation as

\frac{m(t)}{m_0}=(\frac{1}{2})^{\frac{t}{\tau}}

And the term on the left represents the fraction of the radioisotope left after a certain time t.

Therefore, after t = 1 days, the fraction of radioisotope left in the body is

\frac{m(1)}{m_0}=(\frac{1}{2})^{\frac{1}{\tau}}

where the half-life \tau must be expressed in days in order to match the units.

Learn more about radioactive decay:

brainly.com/question/4207569

brainly.com/question/1695370

#LearnwithBrainly

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Temperature change time base problem. Suppose V = 24 V, I = 0.1 A, for water: mw = 51 gm, cw = 4.18 J/gm ∘K-1, for resistor: mr
nirvana33 [79]

Answer:

t = 444.125 sec

Explanation:

Given data:

V = 24 volt

I  = 0.1 ampere

mass of water mw = 51 gm

cr = 4.18 J/gm degree K^-1

mass of resistor = 8 gm

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Read 2 more answers
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