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3 years ago

What are radioactive isotopes?

Physics

1 answer:

melisa1 [442]3 years ago

8 0

Answer:

Radioactive isotope , also called radioisotope,

radionuclide, or radioactive nuclide, any of

several species of the same chemical element

with different masses whose nuclei are unstable

and dissipate excess energy by spontaneously

emitting radiation in the form of alpha , beta , and

gamma rays .

A brief treatment of radioactive isotopes follows.

For full treatment, see isotope: Radioactive

Radioactiveisotopes.

Every chemical element has one or more

radioactive isotopes. For example, hydrogen, the

lightest element, has three isotopes with mass

numbers 1, 2, and 3. Only hydrogen-3 (tritium ),

however, is a radioactive isotope , the other two

being stable. More than 1,000 radioactive

isotopes of the various elements are known.

Approximately 50 of these are found in nature;

the rest are produced artificially as the direct

products of nuclear reactions or indirectly as the

radioactive descendants of these products.

Radioactive isotopes have many useful

applications. In medicine , for example, cobalt -60

is extensively employed as a radiation source to

arrest the development of cancer. Other

radioactive isotopes are used as tracers for

diagnostic purposes as well as in research on

metabolic processes. When a radioactive isotope

is added in small amounts to comparatively large

quantities of the stable element, it behaves

exactly the same as the ordinary isotope

chemically; it can, however, be traced with a

Geiger counter or other detection device.

Iodine -131 has proved effective in treating

hyperthyroidism. Another medically important

radioactive isotope is carbon-14, which is used in

a breath test to detect the ulcer-causing bacteria

Heliobacter pylori .

Explanation:

#learnwithbrainly

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3 years ago

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vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

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Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

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3 years ago

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Answer:

$\dfrac{d\theta}{dt}=0.038\ rad/s$

Explanation:

Given that

$\dfrac{dx}{dt}= -1\ m/s$

From the diagram

$tan\theta=\dfrac{21}{x}$

By differentiating with time t

$sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}$

When x= 10 m

$tan\theta=\dfrac{21}{10}$

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Now by putting the value in equation

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$sec^264.53^{\circ} \dfrac{d\theta}{dt}=-\dfrac{21}{10^2}\times (-1)$

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Answer:

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