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Lady_Fox [76]
2 years ago
5

A sample contains 36 g of a radioactive isotope. How much radioactive isotope remains in the sample after 3 half-lives?

Physics
2 answers:
LiRa [457]2 years ago
5 0
Answer:
4.5g

Explanation:

No = 36g Let half life is T

T= 3 T n number of half lives = t/T=3

N/No= (1/2)^n

Where, No be the initial amount

N be the amount left

n be the number of half lives

N/36 = (1/2)^3

N/36 = 1/8

N = 36/8 = 4.5g
Gre4nikov [31]2 years ago
3 0

Answer:

<u>Option "C":</u> "4.5 g"

Explanation:

N0 = 36 g, Let half-life is T.

t = 3 T, n is number of half lives = t / T = 3

<u>By using the decay law of radioactivity</u>

N / N0 = (1 / 2)^n

where

"N0" be the "initial amount"

"N" be the "amount left"

"n" be the "number of half-lives"

N / 36 = (1/2)^3

N / 36 = 1 / 8

N = 36 / 8 = 4.5 g

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3 years ago
Consider a well-insulated rigid container with two chambers separated by a membrane. The total volume is 5.0 cubic meters. The f
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Answer:

The Entropy generated by the steam = 2.821 kJ/K

Explanation:

Total volume of container = 5m³

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At the first chamber, temperature of water at saturated liquid is 300°C

From the steam table:

Specific enthalpy of saturated liquid at 300°C , h_{f} = 1344.8 kJ/kg

Specific internal energy of saturated liquid at 300°C, U_{f1} =  1332.7 kJ/kg

For closed system, the first law of thermodynamics state that:

dQ = dw + dU..................(1)

work done for free expansion, dw =0

0 = 0 + dU

dU = 0 , i.e. U₁ = U₂

At the second chamber,

The final pressure, P₂ = 50 kPa

From the steam table, at P₂ = 50 kPa,  U_{f2} = 340.49 kJ/kg

(U_{fg} )_{2} =  2142.7 kJ/kg

Let the dryness fraction at the second chamber = x

U_{2} = U_{f2} + U_{fg2}

U_{2} = 340.49 + x2140.7Since U₁ = U₂

1332.7 = 340.49 + x2140.7

Dryness fraction, x = 0.463

From steam table, the specific volume is, u_{f2} = 0.00103 m^{3} /kg\\

u_{2} = u_{f2} + xu_{fg2}

u_{2} = 0.00103 + 0.463(3.2393)\\u_{2} = 1.5 m^{3} /kg\\

u_{2} = \frac{v_{2} }{m_{2} }

V₂ = 5 m³

1.5 = 5/m₂

m₂ = 3.33 kg

At 300°C S_{1} = S_{f} = 3.2548 kJ/kg-k\\

S_{2} = S_{f2} + xS_{fg2}

From the steam table,

S_{f2} = 1.0912 kJ/kg-k\\S_{fg2} = 6.5019 kJ/kg-k\\S_{2} = 1.0912 + 0.463(6.5019)\\S_{2} = 4.102 kJ/kg-k

Therefore the entropy generated will be :

Entropy = mass* (S₂ - S₁)

Entropy = 3.33* (4.102 - 3.2548)

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Answer:

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