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uysha [10]
3 years ago
14

Use the worked example above to help you solve this problem. The half-life of the radioactive nucleus _(88)^(226)text(Ra) is 1.6

103 yr. If a sample initially contains 4.00 1016 such nuclei, determine the following:________.
(a) the initial activity in curies µCi
(b) the number of radium nuclei remaining after 4.4 103 yr nuclei
(c) the activity at this later time µCi
Physics
1 answer:
vfiekz [6]3 years ago
8 0

Answer:

Explanation:

From the information given:

The half-life t_{1/2} = 1.6103 years

The no. of the initial nuclei N_o = 4.00 \times 10^6

Using the formula:

N = N_o exp(-\lambda t)

where;

decay constant \lambda = \dfrac{In2}{1.6*10^3} y^{-1}

∴

N = N_o exp ( \dfrac{-In2}{1.6*10^3}\times 4.4 \times 10^3)

N = N_o exp (- 1.906154747)

The number of radium nuclei N = 5.94 × 10¹⁵

The initial activityA_o = \lambda N_o

A_o =(\dfrac{In (2)}{1.61\times 10^3  \times 365 \times 24 \times 3600}\times 4.00 \times 10^{16})

A_o =546075.8487 \ Bq

Since;

1 curie = 3.7 × 10¹⁰ Bq

Then;

A_o =\dfrac{546075.8487 }{3.7\times 10^{10}}

A_o = 1.47588 \times 10^{-5}Ci

A_o = 14.7588 \  \mu Ci

c) The activity at a later time is:

=5.94 \times 10^{15}( \dfrac{In (2)}{1.60 \times 10^3 \times 365\times 24 \times 3600})

= 81599.09018 \ Bq \\ \\ = \dfrac{81599.09018}{3.7\times 10^{10}} \ Ci \\ \\ = 2.20538 \times 10^6 \ Ci  \\ \\  = 2.20538  \ \mu Ci

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ankoles [38]

Answer:

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Explanation:

a. In order to calculate the magnetic field in the solenoid you use the following formula:

B=\frac{\mu_o n i}{L}               (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

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L: length of the solenoid = 25.0cm = 0.25m

i: current  = 90.0A

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B=\frac{(4\pi*10^{-7}T/A)(460)(90.0A)}{0.25m}=0.20T

The magnetic field in the solenoid is 0.20T

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u=\frac{B^2}{2\mu_o}=\frac{(0.20T)^2}{2(4\pi*10^{-7}T/A)}=17230.6\frac{J}{m^3}

The energy density is 17230.6 J/m³

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V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:

V=AL

A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2

V=(5.5*10^{-5}m^2)(0.25m)=1.375*10^{-5}m^3

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E=(17230.6J/m^3)(1.375*10^{-5}m^3)=0.236J

The energy contained is 0.236J

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L=\frac{\mu_o N^2 A}{L}=\frac{(4\pi*10^{-7}T/A)(460)^2(5.5*10^{-5}m^2)}{0.25m}\\\\L=5.84*10^{-5}H

The inductance of the solenoid is 5.84*10^-5 H

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Answer:

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Explanation:

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Answer:

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ch4aika [34]

Answer:

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