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xxTIMURxx [149]
2 years ago
10

•A radioactive material A (decay constant λA) decays into a material B (decay constant λB) and then into material C (decay const

ant λC) which is also radioactive. Determine the amount of material C remaining after a time t?
Physics
1 answer:
KiRa [710]2 years ago
6 0

Answer:

 The amount of C remaining after time t is

       N_C__{R}} =N_D  =  (N_0 -N_0 e^{\lambda_A t}) - (N_0 -N_0 e^{-\lambda_A t})e^{-\lambda_B } [e^{-\lambda_C }  ]

Explanation:

We can represent the decay sequence as

      A \to B \to C \to D

The reason we added D is because we are told from the question that C is also radioactive so it has the  tendency to decay

Generally for every decay the remaining radioactive element can be obtained as

     N =  N_0 -N_0 e^{- \lambda t}

Where N is the amount of the remaining radioactive material

            N_0 is the original amount amount of the radioactive material before decay

    and  \lambda is the decay constant

Now for the decay from  A \to B  amount of radioactive element B formed from A after time t can be obtained as

          N_b =  N_0 -N_0 e^{- \lambda_A t}

Where \lambda _A is the decay constant of A

  Now for the decay from  B \to C  amount of radioactive element C formed from A after time t can be obtained as

       N_c  =  (N_0 -N_0 e^{\lambda_A t}) - (N_0 -N_0 e^{\lambda_A t})e^{-\lambda_B t}

Where \lambda _B is the decay constant of B

  Now for the decay from  C \to D  amount of radioactive element D formed from A after time t can be obtained as

       N_C__{R}} =N_D  =  (N_0 -N_0 e^{\lambda_A t}) - (N_0 -N_0 e^{-\lambda_A t})e^{-\lambda_B } [e^{-\lambda_C }  ]

So this amount of  D is the reaming amount of the radioactive material C

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IgorLugansk [536]

Given:

The given value is (1.0004)^{\frac{1}{2}}.

To find:

The value of the given expression by using the Binomial approximation.

Explanation:

We have,

(1.0004)^{\frac{1}{2}}

It can be written as:

(1.0004)^{\frac{1}{2}}=(1+0.0004)^{\frac{1}{2}}

(1.0004)^{\frac{1}{2}}=1+\dfrac{1}{2}\times 0.0004      [\because (1+x)^n=1+nx]

(1.0004)^{\frac{1}{2}}=1+0.0002

(1.0004)^{\frac{1}{2}}=1.0002

Therefore, the approximate value of the given expression is 1.0002.

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2 years ago
An object located near the surface of Earth has a weight of a 245 N
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7 0
2 years ago
For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ
slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

        157 lb = 157 lb \times \frac{1 kg}{2.2046 lb}

                   = 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

         180 \frac{mg}{kg} \times 71.215 kg

         = 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

      0.65% (w/w) = \frac{0.65 g}{100 g}

                           = \frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}

                           = \frac{0.65 g}{100 ml}

Therefore, calculate the volume which contains the amount of caffeine as follows.

   12818.7 mg = 12.8187 g = \frac{12.8187 g}{\frac{0.65 g}{100 ml}}

                       = 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

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3 years ago
Mark all the fermions a) Proton b) Electron c)Anti-top d) Gluon e) Tau Neutrino
aliya0001 [1]

Answer:

(a) Proton, (b) Electron (c) tau neutrino.

Explanation:

Elementary particles are the particles without any sub-structure which means they are not composed of other particles.

The elementary particles are classified into three categories which are discussed below:

(1) Quarks: up, down, top, bottom, strange, and charm.

(2) Leptons: electrons, electron neutrino, muon, muon neutrino, tau, tau neutrino.

(3) Bosons: Guon, photons Z bosons, W bosons, Higgs.

Fermions are particles that follow Fermi-Dirac statistics. Fermions consist of all the quarks and leptons. And fermions can also be categorized in elementary particles such as electrons and protons.

Therefore, the fermions in the given question are proton, electron, and tau neutrino.

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3 years ago
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den301095 [7]

Explanation:

Th electric force between charges is inversely proportional to the square of distance between them. It means,

F\propto \dfrac{1}{r^2}

Initial distance, r₁ = 2 cm

Final distance, r₂ = 0.25 cm

Initial force, F₁ = 1 N    

We need to find the electric force between charges if the new separation of 0.25 cm. So,

\dfrac{F_1}{F_2}=(\dfrac{r_2}{r_1})^2\\\\F_2=\dfrac{F_1r_1^2}{r_2^2}\\\\F_2=\dfrac{1\times 2^2}{(0.25)^2}\\\\F_2=64\ N

So, the new force is 64 N if the separation between charges is 64 N.

7 0
3 years ago
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