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10 days ago

Which list of radioisotopes contains an alpha emitter, a beta emitter, and a positron emitter?

Chemistry

2 answers:

alisha [2.7K]10 days ago

6 0

Answer: The correct option is 3.

Explanation: Radioisotopes that emit alpha-particles are termed alpha-emitters. These isotopes undergo alpha-decay.

Those radioisotopes that emit beta-particles $(_{-1}^0\beta )$ are called beta-emitters. They undergo beta-minus decay, in which a neutron converts to a proton and an electron.

Isotopes that emit positrons $(_{+1}^0\beta )$ are known as positron-emitters, undergoing beta-plus decay where a proton becomes a neutron.

From the options given,

Option 1: All three isotopes undergo beta-minus decay.

Option 2: Cs-137 and Tc-99 undergo beta-minus decay.

Fr-220 undergoes alpha-decay.

Option 3: Kr-85 undergoes beta-minus decay.

$_{36}^{85}\textrm{Kr}\rightarrow _{37}^{85}\textrm{Rb}+_{-1}^0\beta$

Ne-19 undergoes positron decay.

$_{10}^{19}\textrm{Ne}\rightarrow _{9}^{19}\textrm{F}+_{+1}^0\beta$

Rn-222 undergoes alpha decay.

$_{86}^{222}\textrm{Rn}\rightarrow _{84}^{218}\textrm{Po}+_{2}^4\alpha$

Option 4: All three isotopes undergo beta-minus decay processes.

Therefore, the correct choice is 3.

eduard [2.5K]10 days ago

4 0

The correct choice is (3): the list contains β-, β+ and α decay types. For (1) the modes are β-, β- and β-. For (2) they are β-, α and β-. For (4) all three are α decay modes.

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The density of air under ordinary conditions at 25°C is 1.19 g/L. How many kilograms of air are in a room that measures 10.0 ft

alisha [2704]

Answer: The air in the room weighs 37.068 kg

Explanation:

Given dimensions:

Length = 10.0 ft

Width = 11.0 ft

Height = 10.0 ft

The volume of the room (rectangular prism) is calculated using:

$V=lbh$

where l = length, b = breadth, h = height.

Substituting the values,

$V=10.0ft\times 11.0ft\times 10.0ft=1100ft^3=31148.53L$

Using the conversion: $1ft^3=28.3168L$

Next, calculate the mass of the air based on density:

$Density=\frac{Mass}{Volume}$

$1.19g/L=\frac{Mass}{31148.53L}$

$Mass=37066.7507g=37.068kg$

Conversion: 1 kg = 1000 g

Thus, the air mass in the room equals 37.068 kilograms.

3 0

1 month ago

Read 2 more answers

Why did we use a mixture of ethanol and water to perform the reaction between the oil and naoh?

alisha [2704]

Ethanol, with the formula C2H5OH, is also referred to as Ethyl alcohol.

Explanation:

The interaction between oil and sodium hydroxide (NaOH) is recognized as the Saponification process.

A combination of ethanol and water yields a fully homogeneous solution, miscible in all proportions.

We use a mixture of ethanol and water for the Saponification because it prevents the fat from reacting with atmospheric oxygen.

This mixture is advantageous as it exhibits lower polarity than water, aiding in the dissolution of non-polar fats, thereby facilitating reaction with sodium hydroxide.

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1 month ago

a nurse practitioner orders Medrol to be given 1.5 mg/kg of body weight. if a child weighs 72.6lb and the available stock of Med

KiRa [2711]

The amount to administer to the child is 2,469 mL.
To convert to kilograms (kg), the child's weight in pounds (lb) is multiplied by 0.45359237: m(child) = 72.6 · 0.045359237 = 32.93 kg.
To find m(Medrol), the child's mass in kilograms is multiplied by 1.5 mg/kg.
Thus, m(Medrol) = 32.93 kg · 1.5 mg/kg = 49.39 mg.
The concentration of Medrol is d(Medrol) = 20.0 mg/mL.
To find the volume of Medrol needed, use V(Medrol) = m(Medrol) ÷ d(Medrol).
So, V(Medrol) = 49.39 mg ÷ 20 mg/mL = 2,469 mL.

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1 month ago

A magnesium ion, Mg2+, with a charge of 3.2×10−19C and an oxide ion, O2−, with a charge of −3.2×10−19C, are separated by a dista

eduard [2509]

Details:

The equation to calculate work done is defined as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

where, k = proportionality constant = $8.99 \times 10^{9} Jm/C^{2}$

$q_{1} = charge of Mg^{2+} = 3.2 \times 10^{-19} C$

$q_{2} = charge of O_{2-} = -3.2 \times 10^{-19} C$

d = separation distance = 0.45 nm = $0.45 \times 10^{-9} m$

Now we will insert the given values into the formula above to compute the work done as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

= $\frac{-[8.99 \times 10^{9} Jm/C^{2} \times 3.2 \times 10^{-19} C \times -3.2 \times 10^{-19} C]}{0.25 \times 10^{-9} m}$

= $3.68 \times 10^{-18} J$

Thus, we can conclude that the work needed to increase the distance between the two ions to infinity is $3.68 \times 10^{-18} J$ .

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20 days ago

Describe the cause of attraction between molecules of water

lions [2649]

In a water molecule, the sharing of electrons occurs between the oxygen and hydrogen atoms within covalent bonds; however, this sharing is unequal. The oxygen atom holds a stronger pull on the electrons compared to the hydrogen atoms in the bond.

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