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ElenaW [278]
3 years ago
15

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

Chemistry
2 answers:
MArishka [77]3 years ago
6 0

Answer: The correct option is 3.

Explanation: Radioisotopes which emits alpha-particle are known as alpha-emitters. These radioisotopes undergo alpha-decay.

The radioisotopes which emits beta-particle (_{-1}^0\beta ) are known as beta-emitters. These radioisotopes undergo beta-minus decay. In this decay a neutron gets converted to a proton and an electron.

The radioisotopes which emits positron-particle (_{+1}^0\beta ) are known as positron-emitters. These radioisotopes undergo beta-plus decay. In this type of decay a proton gets converted to a neutron.

From the given options,

Option 1: All the three radioisotopes undergoes beta-minus decay.

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

Fr-220 is a radioisotope which undergoes alpha-decay.

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

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

Radioisotope Ne-19 undergoes positron decay.

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

Radioisotope Rn-222 undergoes alpha decay.

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

Option 4: All the three radioisotopes undergoes beta-minus decay processes.

Hence, from the above information, the correct option is 3.

vodka [1.7K]3 years ago
4 0
The answer is (3), they are β-, β+ and α decay mode. For (1), they are β-, β- and β- decay mode. For (2), they are β-, α and β-. For (4), they are α, α and α decay mode.
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Dafna11 [192]

Answer:

A. synthesis

Explanation:

The reaction given is a synthesis reaction in which two substances combines to give a product.

In this reaction:

      4Fe + 3O₂ →  2Fe₂O₃

The reaction involves the formation of a single product from two or more reactants.

The formation of a compound from union of the constituent elements also falls into this category of reaction.

3 0
3 years ago
What is the associated deBroglie wavelength of a H2 molecule moving on one direction with kinetic energy of (3/2 kT) at 30 K
andrey2020 [161]

<u>Answer:</u> The de-Broglie's wavelength of a hydrogen molecule is 3.26\AA

<u>Explanation:</u>

Kinetic energy is the measure of temperature of the system.

The equation used to calculate kinetic energy of a particle follows:

E=\frac{3}{2}kT

where,

E = kinetic energy of the particles  = ?

k = Boltzmann constant  = 1.38\times 10^{-23}J/K

T = temperature of the particle = 30 K

Putting values in above equation, we get:

E=\frac{3}{2}\times 1.38\times 10^{-23}J/K\times 30K\\\\E=6.21\times 10^{-22}J

  • Calculating the mass of 1 molecule of hydrogen gas:

Conversion factor used:  1 kg = 1000 g

1 mole of hydrogen gas has a mass of 2 grams or 2\times 10^{-3}kg  

According to mole concept:

6.022\times 10^{23} number of molecules occupy 1 mole of a gas.

As, 6.022\times 10^{23} number of hydrogen molecules has a mass of 2\times 10^{-3}kg

So, 1 molecule of hydrogen will have a mass of = \frac{2\times 10^{-3}kg}{6.022\times 10^{23}}\times 1=3.32\times 10^{-27}kg

  • To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

\lambda=\frac{h}{\sqrt{2mE_k}}

where,

\lambda = De-Broglie's wavelength = ?

h = Planck's constant = 6.624\times 10^{-34}Js

m = mass of 1 hydrogen molecule = 3.32\times 10^{-27}kg

E_k = kinetic energy of the particle = 6.21\times 10^{-22}J

Putting values in above equation, we get:

\lambda=\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 3.32\times 10^{-27}kg\times 6.21\times 10^{-22}J}}

\lambda=3.26\times 10^{-10}m=3.26\AA    (Conversion factor: 1\AA=10^{-10}m )

Hence, the de-Broglie's wavelength of a hydrogen molecule is 3.26\AA

3 0
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Answer:

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