The half-life of phosphorus-32 is 14.26 days. calculate its decay constant. express the decay constant numerically in inverse da
2 answers:
The decay constant(λ) = <u>0.0486/day</u>
<h3>Further explanation</h3>The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.
Usually radioactive elements have an unstable atomic nucleus.
The main particles are emitted by radioactive elements so that they generally decay are alpha (α), beta (β) and gamma (γ) particles
General formulas used in decay:
T = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
The half-life can be expressed in a decay constant( λ)
The half-life of phosphorus-32 is 14.26 days
Then : t 1/2 = 14.26 days
The identity of the missing nucleus
the process of radioactive decay
nuclear fusion and nuclear fission reactions
the product of radioactive decay
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<h3>Further explanation </h3>The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The value of ° H ° can be calculated from the change in enthalpy of standard formation:
∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
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