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vovangra [49]
3 years ago
12

A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact

ive nuclide is left after 10 half-lives? How long will it take for 10 half-lives to pass for plutonium-239?
Physics
2 answers:
maksim [4K]3 years ago
8 0

Answer:

percentage left = 10 %

241 000 years

Explanation:

The half life, t_{\frac{1}{2} } of plutonium-239 is 24 100 years.

Thus, the number of half lives that result in the decrease of the nucleotide is given by the following equation:

The percentage left will be 10 %

if t_{\frac{1}{2} }  = 24 100, then after 10 half-life, the decay will be:

t_{10 years} = 24 100* 10\\                    = 241 000

ArbitrLikvidat [17]3 years ago
7 0

Answer:

  • 0.09 % of the original radioactive nucllde its left after 10 half-lives
  • It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Explanation:

The equation for radioactive decay its:

N ( t) \ = \ N_0 \ e^{ \ -  \frac{t}{\tau}},

where N(t) its quantity of material at time t, N_0 its the initial quantity of material and \tau its the mean lifetime of the radioactive element.

The half-life t_{\frac{1}{2}} its the time at which the quantity of material its the half of the initial value, so, we can find:

N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}

so:

\ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}

e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}

-  \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )

t_{\frac{1}{2}}\ = \tau ln( 2 )

So, after 10 half-lives, we got:

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  t_{\frac{1}{2}}}{\tau}}

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  \tau \ ln( 2 ) }{\tau}}

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  10 \  \ ln( 2 ) }

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

10 \ * \ 24,110 \ years \ = \ 241,100 \ years

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

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The sunlight that reaches Earth is generated by a transformation of _____.
Georgia [21]

nuclear energy

Explanation:

The sunlight that reaches earth is generated by a transformation of nuclear energy. Radiant energy is a from of energy that reaches the earth from nuclear reactions in the core of the sun.

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  • This process does not conserve energy and mass is converted to energy.
  • The nuclear reaction produces huge energy that goes into the solar system space and lights it up.
  • Therefore, it is a transformation of nuclear energy into radiant energy.
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Learn more;

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6 0
3 years ago
How much work in joules is required to lift a 23 kg box up from the ground to your waist that is 1.0 meters high, carry it 6 met
PSYCHO15rus [73]

Answer:

2682

Explanation:

Work done is given by :

Work = Force x distance

         =  mg x d

So, work done in lifting the box of 23 kg up to my waist of 1 m high is :

W = mg x d

   = 23 x 9.18 x 1

   = 211.14

Now work done carrying the box horizontally 6 meters across the room is

W = mg x d

   = 23 x 9.18 x 6

   = 1266.84

Work done in placing the box on the shelf that is 5.7 m above the ground is

W = mg x d

   = 23 x 9.18 x 5.7

   = 1203.49

So the total work done is = 211.14 + 1266.84 + 1203.49

                                          = 2681.47

                                          = 2682 (rounding off)

5 0
2 years ago
Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of
emmainna [20.7K]

Answer:

5,2

Explanation:

From the question we are told that:

Speed of light C=3.0×10^8 m/s.

Generally the equation for Average Speed is mathematically given by

V_{avg}=\frac{d}{t}

Where

d=Distance between the Earth and the sun

d=1.5*10^11m

Therefore

t=\frac{d}{V_{avg}}

t=\frac{1.5*10^11m}{3.0×10^8 m/s.}

t=5*10^2s

Since m and n is given in the form of

m*10^n

Therefore

m=5 & n=2

5,2

3 0
3 years ago
The heat flux that is applied to one face of a plane wall is q″ = 20 W/m2. The opposite face is exposed to air at temperature 30
Blababa [14]

Answer:

400 W/m^2 and 31℃

Explanation:

The output heat flux q"= 20 W/m^2 (geven)

The output heat flux from.the wall to the air by convection

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q"conv = 20(50-30) = 400 W/m^2

Therefor, this case is unsteady and the wall temperature changes with time till the energy balance exist.

ENERGY BALANCE

The input energy must be equal to the output energy for steady state condition. If not the state will be unstaidy or transient.

2. Its noticed that the output heat flux is not that the I put heat flux, therefore the wall tempers will be decreased till the output heat flux is reduced to the value of the given input heat flux

T steady = T∞ +q"/h

= 30 + 20/20 = 31℃

3 0
3 years ago
What are the factors that affect the resistance of a wire?
777dan777 [17]

1) Length of the wire.

2) Thickness of the wire.

3) Temperature.

4) Type of metal.

Hope this helps!

-Payshence

6 0
3 years ago
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