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Yuki888 [10]
2 years ago
12

Describe 3 uses of radioactive isotopes

Physics
2 answers:
pantera1 [17]2 years ago
8 0
Cancer treatments, Radiography, food irradiation
Dennis_Churaev [7]2 years ago
4 0

Answer:

See below.

Explanation:

Radioactive isotopes are isotopes that are unstable; the nucleus decays spontaneously, giving off detectable particles and energy.  An atom with a certain number of protons and neutrons is known as a nuclide. They can be either radioactive or stable. Carbon-12 is a stable carbon nuclide with six protons and six neutrons, whereas C-14 is a radioactive carbon nuclide with six protons and eight neutrons. C-14 is radioactive due to the extra two neutrons, whereas C-12 (and C-13, for that matter) is stable. Oh, and the fine point here is that the same nuclide can have different energy levels. As a result, barium-137 can exist in both a higher and lower energy state; they are distinct isotopes of the same nuclide. Consider a ball resting on top of a table - it might be on top of the table or on the floor just beneath the table. The ball is in the same area in both scenarios, but if it's on top of the table, it has a greater energy level; if it rolls off and falls to the floor, it loses that energy, just as Ba-137m will radioactively decay to Ba-137 (which is stable). OK, so this touches on what makes an atom radioactive in some way. If the nucleus has more energy than it needs, it will give it away by producing radiation. And the source of that extra energy is a mismatch between the amount of protons (which repel each other) and neutrons (which attract each other) (which tend to bind the nucleus together). The nucleus has excessive electrostatic repulsion if there are too few neutrons; if there are too many neutrons, the nucleus has excessive energy gluing it together. Protons become neutrons or neutrons become protons in beta decays; alpha decays remove two protons and two neutrons from the nucleus; in both situations, the nucleus is closer to the lowest-energy configuration that will survive. Oh, and in certain situations, when a radioactive atom decays to generate a non-radioactive one, the "daughter" atom will remain in an excited state for a brief period of time before giving up its energy by generating a gamma-ray photon. As a result, atoms become radioactive because they contain too much energy. So, depending on what they're employed for, certain atoms are radioactive... That's a vast issue, and I won't be able to get into it in depth here. However, here is a list of some of the applications:

  • Medicine

- Cancer Treatment

- imaging for diagnosis (nuclear medicine)

- diagnosis of infectious diseases (radio-immune assay kits)

  • Research

- Investigate biokinetics (where in the body do drugs or other compounds travel)

- dynamics of cells (where in the cell do specific molecules go)

- Sequencing of DNA (not as common today, but still used at times)

- causing mutations to investigate DNA damage and repair processes

- to examine genetic variability by introducing mutations\

  • Industry

- gauges for process control (thickness of glass, paper, steel, etc.)

- density and moisture content of the soil (for construction sites)

- Measurements of quality assurance

- metalworking (e.g. lead and metals analyzers)

- gauges for tank levels (over- and under-flow prevention)

- non-destructive evaluation (industrial radiography)

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A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy
SashulF [63]

Answer:

5.444\times 10^{-4}

Explanation:

The momentum of the neutron before and after the decay  is the same since there's no external force.

P_{sys}=const\\\\P=mv\\\\K=0.5mv^2

#The neutron is initially at rest, so after the decay:

P_A+P_B=0\\\\P_A=-P_B

#After decay, the proton has +ve direction  with a velocity v_Awhile the electron moves in a negative direction with a velocity v_B

Therefore:

P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B

Let the energy released during the decay be Q:

Q=K_{tot}=K_A+K_B\\\\Q=K_A+0.5m_Bv_B^2\\\\Q=K_A+0.5m_B(\frac{m_A}{m_B})^2v_A^2\\\\\ But \ K_A=0.5m_Av_A^2\\\\\therefore Q=K_A+\frac{m_A}{m_B}K_A=K_A(1+\frac{m_A}{m_B})\\\\=Q=\frac{m_A+m_B}{m_B}K_A\\\\m_A=1836m_B\\\\\frac{K_A}{Q}=\frac{m_B}{1836m_B+m_B}=\frac{1}{1837}\\\\\frac{K_A}{Q}=5.444\times10^{-4}

Hence,Kp/Ktot is 5.444x10^(-4)

4 1
3 years ago
A man hits a ball and provides it with an initial velocity of 5.0 m/s on a rough horizontal surface. Due to the surface the ball
Alex777 [14]

Answer:

a) After 5.19 seconds dog catch the ball.

b)  From the dog's initial position at 20.20 m dog catches the ball.

c) Speed of the ball when dog catches ball = 2.405 m/s

   Speed of the dog when dog catches ball = 7.785 m/s

Explanation:

a) Let the time of catching be t.

   We have equation of motion s = ut + 0.5 at²

   Consider the motion of ball

                Initial velocity, u = 5 m/s

                Acceleration, a = -0.5 m/s²

                Time, t = t

                Substituting

                 s = 5 x t + 0.5 x -0.5 x t²

                 s = 5t - 0.25t²

  Consider the motion of dog

                 Initial velocity, u = 0 m/s

                Acceleration, a = 1.5 m/s²

                Time, t = t

                Substituting

                 s + 1 = 0 x t + 0.5 x 1.5 x t²

                 s = 0.75t²      

If they catch up displacement of dog is 1 m more than displacement of ball.

That is

                5t - 0.25t² + 1 =   0.75t²  

                t² - 5t -1 = 0

                t = 5.19 or t = -0.19(not possible)

So after 5.19 seconds dog catch the ball.

b) Displacement of dog, s = 0.75t²  

                            s = 0.75 x 5.19²

                             s = 20.20 m

    From the dog's initial position at 20.20 m dog catches the ball.

c) We have equation of motion v = u + at

       Consider the motion of ball

                Initial velocity, u = 5 m/s

                Acceleration, a = -0.5 m/s²

                Time, t = 5.19 s

                Substituting

                           v = 5 + -0.5 x 5.19 = 2.405 m/s

                Speed of the ball when dog catches = 2.405 m/s

  Consider the motion of dog

                Initial velocity, u = 0 m/s

                Acceleration, a = 1.5 m/s²

                Time, t = 5.19 s

                Substituting

                           v = 0 + 1.5 x 5.19 = 7.785 m/s

                Speed of the dog when dog catches ball = 7.785 m/s

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