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emmainna [20.7K]
2 years ago
11

What type of decay is illustrated by the equation below?

Physics
1 answer:
nlexa [21]2 years ago
3 0

Answer:

The type of decay illustrated by the equation is a Beta decay.

Option B is correct.

Explanation:

Complete Question

What type of decay is illustrated by the equation below?

²¹⁴₈₃Bi → ²¹⁴₈₄Po + ⁰₋₁e

- alpha decay

- beta decay

- positron emission

- electron capture

Solution

The type of radioactive decay is usually discernable from studying the products and reactants (the parent nucleus/atom, the daughter nucleus and the emitted or absorbed particles) of the radioactive decay.

The changes in atomic and mass numbers is accounted for in the particular particle that that is emitted and subsequently termed the type of radioactive decay that is going on.

For example, this equation represented shows that the Bi atom splits into a Po and a Beta particle. Hence, it is easy and straight forward to see that this radioactive decay is to relaese a Beta particle and the decay is a Beta decay.

(Note that, just as presented in the equation, a Beta particle has a mass number of 0 and an atomic number -1, kind of similar to an electron).

- An alpha decay releases an alpha particle.

- A beta decay releases a Beta particle.

- A positron emission releases a positron.

- An electron capture involves an electron attacking the parent atom/nucleus or the reactant.

Hope this Helps!!!

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A mixture of helium and oxygen is used in scuba diving tanks to help prevent ""the bends"". 46 L helium and 12 L oxygen are comb
marin [14]

Answer

given,

For helium

Volume,V = 46 L

Pressure,P = 1 atm

Temperature,T = 25°C  =  273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₁ = ?

number of moles

we know

P V = n R T

n_1 =\dfrac{46 \times 1}{0.0821\times 298}

  n₁ = 1.89 moles

For oxygen

Volume,V = 12 L

Pressure,P = 1 atm

Temperature,T = 25°C  =  273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₂ = ?

number of moles

we know

P V = n R T

n_2 =\dfrac{12 \times 1}{0.0821\times 298}

  n₂ = 0.49 moles

Total volume of tank = 5 L

temperature of tank = 298 K

Partial pressure of helium

P_1=\dfrac{n_1 R T}{V}

P_1=\dfrac{1.89\times 0.0821\times 298}{5}

     P₁ = 9.25 atm

Partial pressure of oxygen

P_2=\dfrac{n_2 R T}{V}

P_2=\dfrac{0.49\times 0.0821\times 298}{5}

    P₂ = 2.39 atm

total pressure

    P = P₁ + P₂

    P = 9.25 + 2.39

    P = 11.64 atm

8 0
2 years ago
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