Polonium-214 went through a series of radioactive decays to produce a stable isotope of bismuth-210 choose the correct decay series that would produce bismuth-210
A: Polonium-214 undergoes 2 beta decay and 1 gamma emission
B: Polonium-214 undergoes 1 alpha decay and 2 gamma emission
C: Polonium-214 undergoes 2 alpha decay and 1 beta decay
D: Polonium-214 undergoes 1 alpha decay 1 beta decay and 1 gamma emission.

Answers

Answer 1

The correct decay series :

₈₄Po²¹⁴⇒₈₃Bi²¹⁰+₂He⁴+ ₋₁e⁰+ ₀γ⁰

Further explanation

Radioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles,  

alpha α particles ₂He⁴ beta β ₋₁e⁰ particles gamma particles ₀γ⁰ positron particles ₁e⁰ neutron ₀n¹

Decay's reaction :

₈₄Po²¹⁴⇒₈₃Bi²¹⁰+x+y+z

From this decay reaction can be concluded

mass number decreased by 4

atomic number decreased by 1

Then the most suitable are:

Polonium-214 undergoes 1 alpha decay 1 beta decay and 1 gamma emission.

₈₄Po²¹⁴⇒₈₃Bi²¹⁰+₂He⁴+ ₋₁e⁰+ ₀γ⁰

So that the mass number and atomic number are the same both before and after decay


Related Questions

How many joules are required to melt 250 grams of water

Answers

Answer:

i not sure but i searched it up and it said Q=104525J

Explanation:

A 12.0% sucrose solution by mass has a density of 1.05 gem, what mass of sucrose is present in a 32.0-mL sample of this solution?
A) 0.126g
B) 3.66g
C) 4.03g
D) 3.84g
E) 280 g

Answers

Answer:

Option C. 4.03 g

Explanation:

Firstly we analyse data.

12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.

Density is the data that indicates grams of solution in volume of solution.

We need to determine, the volume of solution for the concentration

Density = mass / volume

1.05 g/mL = 100 g / volume

Volume =  100 g / 1.05 g/mL → 95.24 mL

Therefore our 12 g of solute are contained in 95.24 mL

Let's finish this by a rule of three.

95.24 mL contain 12 g of sucrose

Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g

Pro
∆G0=-RT in k Known as​

Answers

Answer:

Gibbs free energy equation

Explanation:

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