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3 years ago

What is the relationship between radioactivity and half life

1 answer:

7 0

In Physics, there is an inverse exponential relationship between the half-life and the radioactivity of a substance. Half-life denotes the period during which a specific substance will decrease by 1/2 through radioactive decay. It can also be described as the period during which a single radioactive atom has a 1/2 probability of decaying some time during that specific duration. The shorter the half-life, the more radioactive that substance is. For example, a radioactive substance 'A' with a half-life of 5 days will decay twice as quickly as substance 'B' with a half-life of 10 days. Radioactive decay is a form of exponential decay. In the example above. If you started out with 100g of substance 'A' (half-life of 5 days) and 100 g of substance 'B' (half-life 10 days), then after 10 days, you would have 50g of substance 'B', and only 25g of substance 'A'. Note that, in life sciences, the term 'half-life' is a pharmacalogical term, and has nothing to do with radioactive decay.

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What are the variables of Boyle's law?

lianna [129]

Boyles law is P1V1=P2V2 where P is pressure and V is volume so the answer is pressure and volume

7 0

3 years ago

Calculate the moles of HCl in 15 mL of a 0.50 M solution.

Verizon [17]

Explanation:

$\tiny\implies Molarity = \dfrac{no. \: of \: moles \: of \: solute \times 1000}{volume \: of \: the \: solution \: (in \: ml)}$

$\tiny\implies 0.50 = \dfrac{no. \: of \: moles \: of \: solute \times 1000}{15}$

$\tiny\implies no. \: of \: moles \: of \: solute \times 1000 = 0.50 \times 15$

$\tiny\implies no. \: of \: moles \: of \: solute \times 1000 = 7.5$

$\tiny\implies no. \: of \: moles \: of \: solute = \dfrac{7.5}{1000}$

$\tiny\implies \bf no. \: of \: moles \: of \: solute = 0.075 \: mol$

4 0

3 years ago

1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.

Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice ) = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

$\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}$

$\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}$

$\Delta \ S = {(1*37.7)In \dfrac{263}{273}$

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0

3 years ago

Sodium fluoroacetate (NaO₂C₂H₂F) is a common poison used in New Zealand to control invasive species, such as rats. It can be pre

NISA [10]

Answer:

-219

Explanation:

1.5(339) - 1.5(485) = -219

7 0

3 years ago

Bismuth can undergo beta decay. Which of the elements listed in the table does a bismuth nucleus become when it undergoes beta d

dybincka [34]

Answer:

Lead

Explanation:

Beschde it lead to that I feel like that should be the right answer

6 0

3 years ago